*LeetCode-Find Peak Element

brute force每个和后面的数字比 找到不增长的那个 o(n)

public class Solution {    public int findPeakElement(int[] nums) {        if ( nums == null || nums.length < 2)            return 0;        for ( int i = 0; i < nums.length - 1; i ++ ){            if ( nums[i+1] < nums[i] )                return i;        }        return nums.length-1;    }}

binary search logn

public class Solution {    public int findPeakElement(int[] nums) {        if ( nums == null || nums.length < 2)            return 0;        int high = nums.length-1;        int low = 0;        while ( low < high ){            int mid = ( low + high ) /2;            int mid2 = mid + 1;            if ( nums[mid2] > nums[mid] ){                low = mid2;            }            else                high = mid;        }         return low;    }}