162. Find Peak Element

标签：script   解法   最小   example   use   nlog   amp   find   索引   

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.

解题思路：此题直接秒的，写完还在怀疑这也算是medium难度，看了答案发现有O(nlogn)的解法，我的解法比较暴力，因为两端都有最小值，所以只要找到下一个数比当前数小的位置，当前数就是极大值。看discuss里面用二分查找，比如当前查找到索引x，则如果nums[x+1]>nums[x]，则在x+1到n-1之间必然存在极大值，否则在0到x之间必然存在一个极大值

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

162. Find Peak Element

标签：script   解法   最小   example   use   nlog   amp   find   索引   

原文地址：http://www.cnblogs.com/tsunami-lj/p/7631532.html