From 7887ac0e172fd0e17c217a4a404cd6b1e05ff5d6 Mon Sep 17 00:00:00 2001
From: YDZ <ydz627@gmail.com>
Date: Sun, 22 Sep 2019 18:42:44 +0800
Subject: [PATCH] =?UTF-8?q?=E6=9B=B4=E6=96=B0=E4=BA=8C=E5=88=86=E6=90=9C?=
 =?UTF-8?q?=E7=B4=A2=20tips?=
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---
 README.md | 123 ++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 123 insertions(+)

diff --git a/README.md b/README.md
index 0da364f2..a4470a79 100644
--- a/README.md
+++ b/README.md
@@ -1800,6 +1800,129 @@ func updateMatrix_BFS(matrix [][]int) [][]int {
 
 ## Binary Search
 
+![]()
+
+- 二分搜索的经典写法。需要注意的三点：
+	1. 循环退出条件，注意是 low <= high，而不是 low < high。
+	2. mid 的取值，mid := low + (high-low)>>1
+	3. low 和 high 的更新。low = mid + 1，high = mid - 1。
+
+```go
+func binarySearchMatrix(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if nums[mid] == target {
+			return mid
+		} else if nums[mid] > target {
+			high = mid - 1
+		} else {
+			low = mid + 1
+		}
+	}
+	return -1
+}
+```
+
+- 二分搜索的变种写法。有 4 个基本变种:
+	1. 查找第一个与 target 相等的元素，时间复杂度 O(logn)
+	2. 查找最后一个与 target 相等的元素，时间复杂度 O(logn)
+	3. 查找第一个大于等于 target 的元素，时间复杂度 O(logn)
+	4. 查找最后一个小于等于 target 的元素，时间复杂度 O(logn)
+
+```go
+// 二分查找第一个与 target 相等的元素，时间复杂度 O(logn)
+func searchFirstEqualElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] > target {
+			high = mid - 1
+		} else if nums[mid] < target {
+			low = mid + 1
+		} else {
+			if (mid == 0) || (nums[mid-1] != target) { // 找到第一个与 target 相等的元素
+				return mid
+			}
+			high = mid - 1
+		}
+	}
+	return -1
+}
+
+// 二分查找最后一个与 target 相等的元素，时间复杂度 O(logn)
+func searchLastEqualElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] > target {
+			high = mid - 1
+		} else if nums[mid] < target {
+			low = mid + 1
+		} else {
+			if (mid == len(nums)-1) || (nums[mid+1] != target) { // 找到最后一个与 target 相等的元素
+				return mid
+			}
+			low = mid + 1
+		}
+	}
+	return -1
+}
+
+// 二分查找第一个大于等于 target 的元素，时间复杂度 O(logn)
+func searchFirstGreaterElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] >= target {
+			if (mid == 0) || (nums[mid-1] < 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
+}
+
+// 二分查找最后一个小于等于 target 的元素，时间复杂度 O(logn)
+func searchLastLessElement(nums []int, target int) int {
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + ((high - low) >> 1)
+		if nums[mid] <= target {
+			if (mid == len(nums)-1) || (nums[mid+1] > target) { // 找到最后一个小于等于 target 的元素
+				return mid
+			}
+			low = mid + 1
+		} else {
+			high = mid - 1
+		}
+	}
+	return -1
+}
+```
+
+- 在基本有序的数组中用二分搜索。经典解法可以解，变种写法也可以写，常见的题型，在山峰数组中找山峰，在旋转有序数组中找分界点。第 33 题，第 81 题，第 153 题，第 154 题，第 162 题，第 852 题
+
+```go
+func peakIndexInMountainArray(A []int) int {
+	low, high := 0, len(A)-1
+	for low < high {
+		mid := low + (high-low)>>1
+		// 如果 mid 较大，则左侧存在峰值，high = m，如果 mid + 1 较大，则右侧存在峰值，low = mid + 1
+		if A[mid] > A[mid+1] {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
+```
+
+
 | Title | Solution | Difficulty | Time | Space |收藏| 
 | ----- | :--------: | :----------: | :----: | :-----: | :-----: |
 |[50. Pow(x, n)](https://leetcode.com/problems/powx-n)| [Go](https://github.com/halfrost/LeetCode-Go/tree/master/Algorithms/0050.%20Pow(x%2C%20n))| Medium | O(log n)| O(1)||