From c33b15eeeb389f3af7a7796b9c3d19aa6b352e62 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <tangshusen@pku.edu.cn>
Date: Tue, 27 Aug 2019 22:00:21 +0800
Subject: [PATCH] Create 162. Find Peak Element.md

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+# [162. Find Peak Element](https://leetcode.com/problems/find-peak-element/)
+# 思路
+寻找一个数组中的极大值. 由于要求对数复杂度, 所以我们很容易想到二分法. 
+由于题目说了nums[-1]和nums[n]为负无穷, 那么极大值一定存在(想象一下被山谷包围的山脉). 
+
+那么在二分的循环中:
+1. 若`mid == 0 || nums[mid-1] < nums[mid]`, 即mid左边是上升趋势, 那么mid或者其右边一定存在极大值;
+2. 满足1且`mid == n - 1 || nums[mid] > nums[mid+1]`, 即mid就是极大值;
+3. 满足1但不满足2, 那么其右边一定存在极大值, 此时更新`low = mid + 1`;
+4. 否则, mid左边一定存在极大值, 此时更新`high = mid - 1`;
+
+# C++
+``` C++
+class Solution {
+public:
+    int findPeakElement(vector<int>& nums) {
+        int n = nums.size();
+        int low = 0, high = n - 1, mid;
+        while(low < high){
+            mid = low + (high - low) / 2;
+            if(mid == 0 || nums[mid-1] < nums[mid]){
+                if(mid == n - 1 || nums[mid] > nums[mid+1]) return mid; // 找到
+                else low = mid + 1;
+            }
+            else high = mid - 1;
+        }
+        return low;
+    }
+};
+```