添加 problem 436

Author: halfrost

Algorithms/436. Find Right Interval/436. Find Right Interval.go

@@ -0,0 +1,73 @@
+package leetcode
+
+import "sort"
+
+// 解法一 利用系统函数 sort + 二分搜索
+func findRightInterval(intervals [][]int) []int {
+	intervalList := make(intervalList, len(intervals))
+	// 转换成 interval 类型
+	for i, v := range intervals {
+		intervalList[i] = interval{interval: v, index: i}
+	}
+	sort.Sort(intervalList)
+	out := make([]int, len(intervalList))
+	for i := 0; i < len(intervalList); i++ {
+		index := sort.Search(len(intervalList), func(p int) bool { return intervalList[p].interval[0] >= intervalList[i].interval[1] })
+		if index == len(intervalList) {
+			out[intervalList[i].index] = -1
+		} else {
+			out[intervalList[i].index] = intervalList[index].index
+		}
+	}
+	return out
+}
+
+type interval struct {
+	interval []int
+	index    int
+}
+
+type intervalList []interval
+
+func (in intervalList) Len() int { return len(in) }
+func (in intervalList) Less(i, j int) bool {
+	return in[i].interval[0] <= in[j].interval[0]
+}
+func (in intervalList) Swap(i, j int) { in[i], in[j] = in[j], in[i] }
+
+// 解法二 手撸 sort + 二分搜索
+func findRightInterval1(intervals [][]int) []int {
+	if len(intervals) == 0 {
+		return []int{}
+	}
+	intervalsList, res, intervalMap := []Interval{}, []int{}, map[Interval]int{}
+	for k, v := range intervals {
+		intervalsList = append(intervalsList, Interval{Start: v[0], End: v[1]})
+		intervalMap[Interval{Start: v[0], End: v[1]}] = k
+	}
+	quickSort(intervalsList, 0, len(intervalsList)-1)
+	for _, v := range intervals {
+		tmp := searchFirstGreaterInterval(intervalsList, v[1])
+		if tmp > 0 {
+			tmp = intervalMap[intervalsList[tmp]]
+		}
+		res = append(res, tmp)
+	}
+	return res
+}
+
+func searchFirstGreaterInterval(nums []Interval, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid].Start >= target {
+			if (mid == 0) || (nums[mid-1].Start < target) { // 找到第一个大于等于 target 的元素
+				return mid
+			}
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return -1
+}

Algorithms/436. Find Right Interval/436. Find Right Interval_test.go

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+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question436 struct {
+	para436
+	ans436
+}
+
+// para 是参数
+// one 代表第一个参数
+type para436 struct {
+	one [][]int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans436 struct {
+	one []int
+}
+
+func Test_Problem436(t *testing.T) {
+
+	qs := []question436{
+
+		question436{
+			para436{[][]int{[]int{3, 4}, []int{2, 3}, []int{1, 2}}},
+			ans436{[]int{-1, 0, 1}},
+		},
+
+		question436{
+			para436{[][]int{[]int{1, 4}, []int{2, 3}, []int{3, 4}}},
+			ans436{[]int{-1, 2, -1}},
+		},
+
+		question436{
+			para436{[][]int{[]int{1, 2}}},
+			ans436{[]int{-1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 436------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans436, q.para436
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findRightInterval(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}

Algorithms/436. Find Right Interval/README.md

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+# [436. Find Right Interval](https://leetcode.com/problems/find-right-interval/)
+
+
+## 题目:
+
+Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i.
+
+For any interval i, you need to store the minimum interval j's index, which means that the interval j has the minimum start point to build the "right" relationship for interval i. If the interval j doesn't exist, store -1 for the interval i. Finally, you need output the stored value of each interval as an array.
+
+**Note:**
+
+1. You may assume the interval's end point is always bigger than its start point.
+2. You may assume none of these intervals have the same start point.
+
+**Example 1:**
+
+    Input: [ [1,2] ]
+    
+    Output: [-1]
+    
+    Explanation: There is only one interval in the collection, so it outputs -1.
+
+**Example 2:**
+
+    Input: [ [3,4], [2,3], [1,2] ]
+    
+    Output: [-1, 0, 1]
+    
+    Explanation: There is no satisfied "right" interval for [3,4].
+    For [2,3], the interval [3,4] has minimum-"right" start point;
+    For [1,2], the interval [2,3] has minimum-"right" start point.
+
+**Example 3:**
+
+    Input: [ [1,4], [2,3], [3,4] ]
+    
+    Output: [-1, 2, -1]
+    
+    Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
+    For [2,3], the interval [3,4] has minimum-"right" start point.
+
+**NOTE:** input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
+
+
+## 题目大意
+
+给定一组区间，对于每一个区间 i，检查是否存在一个区间 j，它的起始点大于或等于区间 i 的终点，这可以称为 j 在 i 的“右侧”。
+
+对于任何区间，你需要存储的满足条件的区间 j 的最小索引，这意味着区间 j 有最小的起始点可以使其成为“右侧”区间。如果区间 j 不存在，则将区间 i 存储为 -1。最后，你需要输出一个值为存储的区间值的数组。
+
+注意:
+
+- 你可以假设区间的终点总是大于它的起始点。
+- 你可以假定这些区间都不具有相同的起始点。
+
+
+## 解题思路
+
+
+- 给出一个 `interval` 的 数组，要求找到每个 `interval` 在它右边第一个 `interval` 的下标。A 区间在 B 区间的右边：A 区间的左边界的值大于等于 B 区间的右边界。
+- 这一题很明显可以用二分搜索来解答。先将 `interval` 数组排序，然后针对每个 `interval`，用二分搜索搜索大于等于 `interval` 右边界值的 `interval`。如果找到就把下标存入最终数组中，如果没有找到，把 `-1` 存入最终数组中。