【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.

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Note:

Your solution should be in logarithmic complexity.

 

這題就是求序列最大值。順序查找或二分查找都可。

知足複雜度要求的話須要用二分查找。

 

解法一：順序查找

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

 

解法二：二分查找（遞歸）

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        return Helper(nums, 0, nums.size()-1);
    }
    int Helper(vector<int>& nums, int low, int high)
    {
        if(low == high)
            return low;
        int mid = low + (high-low)/2;
        if(nums[mid] > nums[mid+1])
            return Helper(nums, low, mid);
        else
            return Helper(nums, mid+1, high);
    }
};  

 

解法三：二分查找（迭代）

class Solution {
public:
    int findPeakElement(vector<int>& nums) {
        int low = 0;
        int high = nums.size()-1;
        while(low < high)
        {
            int mid = low + (high-low)/2;
            if(nums[mid] > nums[mid+1])
                high = mid;
            else
                low = mid+1;
        }
        return low;
    }
};