leetcode-162-Find Peak Element

                                                Find Peak Element

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.

click to show spoilers.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

输入任意一个比相邻两元素都大的那个数的下标。

比较2n次

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        for(int i=0;i<nums.size();i++){
            if(i==0&&nums[i]>nums[i+1]) return i;
            if(i==nums.size()-1&&nums[i]>nums[i-1]) return i;
            if(nums[i]>nums[i-1]&&nums[i]>nums[i+1]) return i;
        }
    }
};

比较n次。比后一个元素大即可，因为如果该元素比前一个小，就输出了。

int findPeakElement(int* nums, int numsSize) {
    for(int i=0;i<numsSize-1;i++){
        if(nums[i]>nums[i+1]) return i;
    }
    return numsSize-1;
}

leetcode-162-Find Peak Element

原文：http://blog.csdn.net/u014705854/article/details/46794755