From f3cddc5118b8d410a6dc262f5c121ed60142468b Mon Sep 17 00:00:00 2001
From: Bugsource <35368347+Bugsource@users.noreply.github.com>
Date: Tue, 30 Jul 2019 17:54:39 +0800
Subject: [PATCH] Create 162. Find Peak Element.md

---
 BinarySearch/162. Find Peak Element.md | 60 ++++++++++++++++++++++++++
 1 file changed, 60 insertions(+)
 create mode 100644 BinarySearch/162. Find Peak Element.md

diff --git a/BinarySearch/162. Find Peak Element.md b/BinarySearch/162. Find Peak Element.md
new file mode 100644
index 0000000..4c53946
--- /dev/null
+++ b/BinarySearch/162. Find Peak Element.md	
@@ -0,0 +1,60 @@
+# 题目描述
+```
+A peak element is an element that is greater than its neighbors.
+
+Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index.
+
+The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
+
+You may imagine that nums[-1] = nums[n] = -∞.
+
+Example 1:
+
+Input: nums = [1,2,3,1]
+Output: 2
+Explanation: 3 is a peak element and your function should return the index number 2.
+Example 2:
+
+Input: nums = [1,2,1,3,5,6,4]
+Output: 1 or 5 
+Explanation: Your function can return either index number 1 where the peak element is 2, 
+             or index number 5 where the peak element is 6.
+```
+
+# 解法
+此题就是考数学，如果一个区间[a,b], dfa/dx > 0, dfb/dx < 0， 则[a,b]上必有导数为0的点，即为极值点。
+## 解法1
+暴力扫描，不过注意到-1到0为递增，所以只要找到第一个递减的转折点，否则就是数组最末尾元素
+```
+class Solution {
+    public int findPeakElement(int[] nums) {
+        int preDirection = 1, len = nums.length;
+        for(int i = 1; i < len; i++){
+            if(nums[i] < nums[i-1]){
+                return i-1;
+                
+            }
+        }        
+        return len-1;//如果能执行到这里，说明整个数组就是递增的        
+    }
+}
+```
+
+## 解法2
+二分缩小查找范围，缩小依据即上文提到的数学原理
+```
+class Solution {
+    public int findPeakElement(int[] nums) {
+        int low = 0, high = nums.length-1;
+        while(low < high){
+            int mid = low + (high-low)/2;
+            if(nums[mid] > nums[mid+1]){//high>=mid+1，所以mid+1不会越界
+                high = mid;//查找区间缩小为[0,mid]
+            }else{
+                low = mid+1;//[mid+1,high]
+            }
+        }
+        return low;      
+    }
+}
+```