Find Peak Element：寻找数列中的局部峰值

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.

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Note:

Your solution should be in logarithmic complexity.

思路一：从头遍历，寻找峰值。复杂度O(n)

class Solution {    public int findPeakElement(int[] nums) {        if(nums.length==0) return -1;        if(nums.length==1) return 0;        Stack<Integer> s = new Stack<Integer>();        for(int i = 0;i<nums.length;i++){            if(s.isEmpty()){                s.push(nums[i]);            }else{                int temp = s.peek();                if(temp<nums[i]){                    s.push(nums[i]);                }else{                    return i-1;                }            }        }        return nums.length-1;    }}

思路二：这类的题就不能从头遍历，可以考虑二分查找。既然是局部峰值，那么查找局部最大值即可。复杂度O(logN)

class Solution {    public int findPeakElement(int[] nums) {        if(nums.length==0) return -1;        if(nums.length==1) return 0;        int l = 0;        int r = nums.length-1;        while(l<r){            int m1 =(l+r)/2;            int m2 = m1+1;            if(nums[m1]<nums[m2]){                l = m2;            }else{                r = m1;            }        }        return l;    }}