Search for a Range

Author: mengli

Search for a Range.java

@@ -0,0 +1,63 @@
+Given a sorted array of integers, find the starting and ending position of a given target value.
+
+Your algorithm's runtime complexity must be in the order of O(log n).
+
+If the target is not found in the array, return [-1, -1].
+
+For example,
+Given [5, 7, 7, 8, 8, 10] and target value 8,
+return [3, 4].
+
+public class Solution {
+    public int[] searchRange(int[] A, int target) {
+        int low = findLow(A, target, 0, A.length - 1);
+        int high = findHigh(A, target, 0, A.length - 1);
+        int[] ret = new int[2];
+        ret[0] = low;
+        ret[1] = high;
+        return ret;
+    }
+    
+    private int findLow(int[] A, int target, int l, int r) {
+        int mid = 0;
+        int ret = -1;
+        while (l <= r) {
+            mid = (l + r) / 2;
+            if (A[mid] == target) {
+                ret = mid;
+                int next = findLow(A, target, l, mid - 1);
+                if (next != -1) {
+                    ret = next;
+                }
+                break;
+            } else if (A[mid] < target) {
+                l = mid + 1;
+            } else {
+                r = mid - 1;
+            }
+            
+        }
+        return ret;
+    }
+
+    private int findHigh(int[] A, int target, int l, int r) {
+        int mid = 0;
+        int ret = -1;
+        while (l <= r) {
+            mid = (l + r) / 2;
+            if (A[mid] == target) {
+                ret = mid;
+                int next = findHigh(A, target, mid + 1, r);
+                if (next != -1) {
+                    ret = next;
+                }
+                break;
+            } else if (A[mid] < target) {
+                l = mid + 1;
+            } else {
+                r = mid - 1;
+            }
+        }
+        return ret;
+    }
+}