【LeetCode】162. Find Peak Element (3 solutions)

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.

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.

Note:

Your solution should be in logarithmic complexity.

这题就是求序列最大值。顺序查找或二分查找均可。

满足复杂度要求的话需要用二分查找。

解法一：顺序查找

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

解法二：二分查找（递归）

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        return Helper(nums, , nums.size()-);
    }
    int Helper(vector<int>& nums, int low, int high)
    {
        if(low == high)
            return low;
        int mid = low + (high-low)/;
        if(nums[mid] > nums[mid+])
            return Helper(nums, low, mid);
        else
            return Helper(nums, mid+, high);
    }
};  

解法三：二分查找（迭代）

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int low = ;
        int high = nums.size()-;
        while(low < high)
        {
            int mid = low + (high-low)/;
            if(nums[mid] > nums[mid+])
                high = mid;
            else
                low = mid+;
        }
        return low;
    }
};