LeetCode: 162. Find Peak Element

又是一道加了锁，不能做的题。
只好做 162 咯。

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.

题目大意：找出给定数组中大于相邻的元素的元素下标。

解题思路

遍历一遍，判断当前元素和相邻元素的大小关系。

AC 代码

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        for(size_t i = 0; i < nums.size(); ++i)
        {
            int leftFlag = false, rightFlag = false;
            if(i == 0 || nums[i-1] < nums[i]) leftFlag = true;
            if(i+1 == nums.size() || nums[i+1] < nums[i]) rightFlag = true;

            if(leftFlag && rightFlag) return i;
        }

        return -1;
    }
};