【leetcode】Array——Find Peak Element（162）

题目：

A peak element is an element that is greater than its neighbors.

Given an input array where num[i] ≠ num[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 num[-1] = num[n] = -∞.

For example, in array [1, 2, 3, 1], 3 is a peak element and your function should return the index number 2.

Your solution should be in logarithmic complexity.

思路：要求时间复杂度O(logN)，立刻就想到了Binary Search

取mid，然后判断nums[mid-1] nums[mid] nums[mid+1]之间的大小，判断mid位于巅峰的左侧还是右侧。

要注意的细节，因为mid可能会是0，导致nums[mid-1]越界，所以要处理

代码：

    public int findPeakElement(int[] nums) {    int left=0;    int right=nums.length-1;    while(left<right){    //亦可以提前结束循环，这样不会出现mid=0的情况//    if(left+1==right)//    return (nums[left]>nums[right])?left:right;    int mid = left+((right-left)>>1);    if(mid==0){//当出现mid=0要特殊处理，不然mid-1会越界    return (nums[0]>nums[1])?0:1;    }    if(nums[mid]>nums[mid-1]&&nums[mid]>nums[mid+1])    return mid;    if(nums[mid-1]<nums[mid+1]){//上升阶段    left=mid+1;    }else{//下降阶段    right=mid-1;    }    }        return left;    }