【LeetCode】162. Find Peak Element (3 solutions)-优快云博客

本文详细介绍了如何使用顺序查找和二分查找两种方法来解决寻找序列中最大值的问题，并通过实例代码展示了实现过程。适用于对算法效率有要求的情景。

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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.

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 == 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（递归）

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;
    }
};