162. Find Peak Element

标签：==   HERE   BMI   fast   number   and   pre   tor   run   

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.

Note:

Your solution should be in logarithmic complexity.

Approach #1:

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int len = nums.size();
        if (len == 1) return 0;
        int start = 1, end = len-2;
        while (nums[0] == nums[start] && start < len) start++;
        if (nums[0] > nums[start]) return 0;
        while (nums[len-1] == nums[end] && end >= 0) end--;
        if (nums[len-1] > nums[end]) return len-1;
        for (int i = start; i <= end; ++i) {
            if (nums[i] > nums[i-1] && nums[i] > nums[i+1]) return i;
        }
    }
};

162. Find Peak Element

标签：==   HERE   BMI   fast   number   and   pre   tor   run   

原文地址：https://www.cnblogs.com/ruruozhenhao/p/9902410.html