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menglimengli
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Write some code.
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5 files changed

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‎Interleaving String.java‎

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Original file line numberDiff line numberDiff line change
@@ -23,4 +23,51 @@ public boolean isInterleave(String s1, String s2, String s3) {
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mat[i][j] = (mat[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1)) || (mat[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1));
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return mat[s1.length()][s2.length()];
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}
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}
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public class Solution {
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public boolean isInterleave(String s1, String s2, String s3) {
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return search(s1, s2, s3, 0, 0, 0);
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}
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private boolean search(String s1, String s2, String s3, int i1, int i2, int i3) {
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int len1 = s1.length(), len2 = s2.length();
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if (i1 < len1 && i2 < len2) {
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boolean result = false;
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if (s1.charAt(i1) == s3.charAt(i3)) {
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i1++;
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i3++;
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result = search(s1, s2, s3, i1, i2, i3);
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if (result) {
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return true;
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} else {
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i1--;
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i3--;
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}
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}
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if (s2.charAt(i2) == s3.charAt(i3)) {
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i2++;
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i3++;
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result = search(s1, s2, s3, i1, i2, i3);
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return result;
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}
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return false;
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}
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if (i1 < len1) {
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return s1.substring(i1).compareTo(s3.substring(i3)) == 0;
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}
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return s2.substring(i2).compareTo(s3.substring(i3)) == 0;
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}
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}

‎Maximal Rectangle.java‎

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Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area.
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public class Solution {
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public int maximalRectangle(char[][] matrix) {
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int row = matrix.length;
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if (row == 0) return 0;
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int col = matrix[0].length;
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int[][] map = new int[row][col];
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int area = 0;
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for (int i = 0; i < row; i++) {
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for (int j = 0; j < col; j++) {
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int prev = 0;
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if (i != 0) {
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prev = map[i - 1][j];
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}
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if (matrix[i][j] == '1') {
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map[i][j] = prev + 1;
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} else {
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map[i][j] = prev;
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}
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}
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}
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for (int i = 0; i < row; i++) {
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for (int j = i; j < row; j++) {
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int[] line = new int[col];
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for (int k = 0; k < col; k++) {
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line[k] = map[j][k] - map[i][k] + (matrix[i][k] == '0' ? 0 : 1);
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}
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int l = 0;
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int tmp = 0;
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for (int f = 0; f < col; f++) {
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if (line[f] == j - i + 1) tmp++;
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else {
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if (tmp > l) {
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l = tmp;
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}
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tmp = 0;
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}
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}
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if (tmp > l) {
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l = tmp;
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}
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int s = (j - i + 1) * l;
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if (s > area) {
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area = s;
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}
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}
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}
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return area;
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}
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}

‎Scramble String.java‎

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Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.
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Below is one possible representation of s1 = "great":
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great
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/ \
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gr eat
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/ \ / \
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g r e at
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/ \
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a t
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To scramble the string, we may choose any non-leaf node and swap its two children.
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For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".
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rgeat
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/ \
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rg eat
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/ \ / \
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r g e at
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/ \
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a t
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We say that "rgeat" is a scrambled string of "great".
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Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".
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rgtae
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/ \
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rg tae
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/ \ / \
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r g ta e
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/ \
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t a
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We say that "rgtae" is a scrambled string of "great".
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Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.
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public class Solution {
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public boolean isScramble(String s1, String s2) {
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int length1 = s1.length();
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int length2 = s2.length();
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if (length1 != length2)
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return false;
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if (length1 == 0 || s1.equals(s2))
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return true;
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char[] ca1 = s1.toCharArray();
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char[] ca2 = s2.toCharArray();
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Arrays.sort(ca1);
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Arrays.sort(ca2);
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if (!Arrays.equals(ca1, ca2))
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return false;
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int i = 1;
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while (i < length1) {
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String a1 = s1.substring(0, i);
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String b1 = s1.substring(i, length1);
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String a2 = s2.substring(0, i);
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String b2 = s2.substring(i, length2);
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if (a1.equals(b2) && b1.equals(a2)) return true;
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boolean r = isScramble(a1, a2) && isScramble(b1, b2);
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if (!r) {
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String c2 = s2.substring(0, length1 - i);
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String d2 = s2.substring(length1 - i);
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r = isScramble(a1, d2) && isScramble(b1, c2);
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}
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if (r) return true;
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i++;
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}
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return false;
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}
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}

‎Set Matrix Zeroes.java‎

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Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
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Follow up:
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Did you use extra space?
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A straight forward solution using O(mn) space is probably a bad idea.
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A simple improvement uses O(m + n) space, but still not the best solution.
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Could you devise a constant space solution?
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public class Solution {
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public void setZeroes(int[][] matrix) {
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int row = matrix.length;
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if (row == 0) return;
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int col = matrix[0].length;
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for (int i = 0; i < row; i++) {
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for (int j = 0; j < col; j++) {
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if (matrix[i][j] == 0) {
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for (int k = 0; k < col; k++) {
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if (matrix[i][k] != 0)
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matrix[i][k] = -1;
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}
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for (int f = 0; f < row; f++) {
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if (matrix[f][j] != 0)
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matrix[f][j] = -1;
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}
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}
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}
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}
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for (int i = 0; i < row; i++) {
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for (int j = 0; j < col; j++) {
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if (matrix[i][j] == -1) {
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matrix[i][j] = 0;
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}
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}
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}
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}
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}

‎Unique Binary Search Trees.java‎

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Given n, how many structurally unique BST's (binary search trees) that store values 1...n?
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For example,
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Given n = 3, there are a total of 5 unique BST's.
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1 3 3 2 1
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\ / / / \ \
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3 2 1 1 3 2
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/ / \ \
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2 1 2 3
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public class Solution {
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public int numTrees(int n) {
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if (n == 1) return 1;
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if (n == 2) return 2;
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int[] record = new int[n + 1];
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record[0] = 1;
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record[1] = 1;
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record[2] = 2;
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for (int i = 3; i <= n; i++) {
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int tmp = 0;
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for (int k = 0; k < i; k++) {
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tmp += (record[k] * record[i - k - 1]);
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}
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record[i] = tmp;
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}
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return record[n];
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}
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}

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