LeetCode 845——数组中的最长山脉

1. 题目

2. 解答

2.1 方法一

left 数组表示当前元素左边比当前元素小的元素个数，right 数组数组表示当前元素右边比当前元素小的元素个数。在山脉的中间 B[i] 处，其左边和右边肯定都有小于 B[i] 的元素，而山脉的长度即为 left[i] + right[i] + 1。

class Solution {
public:
    int longestMountain(vector<int>& A) {
        
        int n = A.size();
        if (n < 3)    return 0;
        
        vector<int> left(n, 0);
        vector<int> right(n, 0);
        
        for (int i = 1; i < n; i++)
        {
            if (A[i] > A[i-1])  left[i] = left[i-1] + 1;
        }
        
        for (int i = n-2; i >= 0; i--)
        {
            if (A[i] > A[i+1])  right[i] = right[i+1] + 1;;
        }
        
        int len = 0;
        for (int i = 0; i < n; i++)
        {
            if (left[i] != 0 && right[i] != 0)
                len = max(len, left[i] + right[i] + 1);
        }
        
        return len;   
    }
};

2.2 方法二

max_than_left 数组若为 1 则表明当前元素比左边元素大，max_than_right 数组若为 1 则表明当前元素比右边元素大。

若为山脉则两个数组应该为如下序列

数组取值
max_than_left	1(可选)	...	1(必选)	...	0(可选)
max_than_right	0(可选)	...	1(必选)	...	1(可选)

class Solution {
public:
    int longestMountain(vector<int>& A) {
        
        
        int n = A.size();
        if (n < 3)    return 0;
        
        vector<int> max_than_left(n, 0);
        vector<int> max_than_right(n, 0);
        
        for (int i = 1; i < n; i++)
        {
            if (A[i] > A[i-1])  max_than_left[i] = 1;
        }
        
        for (int i = 0; i < n-1; i++)
        {
            if (A[i] > A[i+1])  max_than_right[i] = 1;
        }
        
        int result = 0;
        int len = 0;
        int left_flag = 0;
        int middle_flag = 0;
        int right_flag = 0;
        
        for (int i = 0; i < n; i++)
        {
            if (max_than_left[i] == 1 && max_than_right[i] == 0)
            {
                if (left_flag)  result++;    
                else    
                {
                    result = 3;
                    left_flag = 1;
                }   
            }
            else if (max_than_left[i] == 1 && max_than_right[i] == 1)
            {
                if (left_flag)  
                {
                    result++;  
                }
                else
                {
                    result = 3;
                }
                
                left_flag = 0;
                middle_flag = 1;
            }
            else if (max_than_left[i] == 0 && max_than_right[i] == 1)
            {
                if (right_flag) result++;
                if (middle_flag)     
                {
                    middle_flag = 0;
                    right_flag = 1;
                    result++;
                }  
            }
            else
            {
                right_flag = 0;
                middle_flag = 0;
                left_flag = 0;
                result = 0;
            }
            if (middle_flag || right_flag) len = max(len, result);
        }
        
        return len;
        
    }
};

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