162. Find Peak Element - jiejackyzhang/leetcode-note GitHub Wiki

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.

Note: Your solution should be in logarithmic complexity.

解题思路为binary search。

有如下几种情形：

1. nums[mid-1] < nums[mid] > nums[mid+1]，则mid是peak，返回结果；
2. nums[mid-1] < nums[mid] < nums[mid+1]，则peak在mid右边；
3. nums[mid-1] > nums[mid] > nums[mid+1]，则peak在mid左边；
4. nums[mid-1] > nums[mid] < nums[mid+1]，则mid左右两边都存在peak，任取一边即可。

public class Solution {
    public int findPeakElement(int[] nums) {
        if(nums == null || nums.length == 0) return -1;
        int start = 0, end = nums.length-1;
        while(start + 1 < end) {
            // at least 3 elements
            int mid = start + (end - start) / 2;
            if(nums[mid-1] < nums[mid] && nums[mid] > nums[mid+1]) {
                return mid;
            } else if(nums[mid-1] < nums[mid] && nums[mid] < nums[mid+1]) {
                start = mid;
            } else {
                end = mid;
            }
        }
        return nums[start] > nums[end] ? start : end;
    }
}