LeetCode——42.接雨水【前后缀 & 双指针】

题解

• 方法一：暴力的优化，前后缀数组预处理
  
• 方法二：双指针
  

AC-Code

• 前后缀数组

class Solution {
public:
    int trap(vector<int>& height) {
        int n = height.size();
        
        int* pre = new int[n];
        int* suf = new int[n];
        memset(pre, 0, sizeof(int) * n);
        memset(suf, 0, sizeof(int) * n);
        
        for(int i = 0, maxHeight = 0; i < n; ++i)
            pre[i] = maxHeight = max(maxHeight, height[i]);
        
        for(int i = n - 1, maxHeight = 0; i > 0; --i)
            suf[i] = maxHeight = max(maxHeight, height[i]);
        
        int result = 0;
        for(int i = 0; i < n; ++i){
            result += min(pre[i], suf[i]) - height[i] > 0 ?
                min(pre[i], suf[i]) - height[i] : 0;
        }
        
        return result;
    }
};

• 双指针

class Solution {
public:
    int trap(vector<int>& height) {
        int n = height.size();
        
        int result = 0;
        int left = 0, right = n - 1;
        int leftMax = 0, rightMax = 0;
        
        while(left <= right) {	// 不要少判断最中间的位置
            if(leftMax < rightMax) {	// 如果leftMax小，那么leftMax可信，利用它可以求left可以存储的rain water
                leftMax > height[left] ? 
                    result += leftMax - height[left] : (leftMax = height[left]);
                ++left;
            }
            else {
                rightMax > height[right] ?
                    result += rightMax - height[right] : (rightMax = height[right]);
                --right;
            }
        }
        
        return result;
    }
};

标签：pre,后缀,42,maxHeight,height,int,result,LeetCode,left
来源： https://blog.csdn.net/Q_1849805767/article/details/105315895