【22】42. Trapping Rain Water

42. Trapping Rain Water (cf.container with most water)

• Total Accepted: 98539
• Total Submissions: 277383
• Difficulty: Hard
• Contributors: Admin

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

For example, 
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

 Solution 1：

class Solution {
public:
    int trap(vector<int>& height) {
        int n = height.size();
        if(n < 2) return 0;
        
        vector<int> maxL(n, 0), maxR(n, 0);//问题出现在这里 vector要初始化 或者用pushback加值
        
        int max = height[0];
        //从左往右扫，记录每一个index的左边最高是多少
        
        maxL[0] = 0;
        //for(int i = 0; i < n; i++){
        for(int i = 1; i < n - 1; i++){
            maxL[i] = max;
            if(max < height[i]){
                max = height[i];
            }
        }
        //从右往左扫，记录每一个index的右边的最高是多少
        //从右往左扫，记录最大圈水数
        max = height[n - 1];
        int trap = 0;
        int traps = 0;
        
        maxR[n - 1] = 0;
        //for(int i = n - 1; i >= 0; i--){
        for(int i = n - 2; i > 0; i--){
            maxR[i] = max;
            trap = min(maxL[i], maxR[i]) - height[i];
            if(trap > 0){
                traps += trap;
            }
            if(max < height[i]){
                max = height[i];
            }
        }
        return traps;
    }
};

Solution 2：两边向中间扫（left，right）

只需遍历一次。left, right指针分别指向头尾。如果left对应的height值小于right，则向右扫；反之向左。如果当前值小于left与right比较的较小值，则把差值存入结果。

class Solution {
public:
    int trap(vector<int>& height) {
        int res = 0;
        int left = 0; 
        int right = height.size() - 1;
        while(left < right){
            int mn = min(height[left], height[right]);
            if(mn == height[left]){
                left++;
                while(height[left] < mn && left < right){
                    res += mn - height[left++];//别忘了++
                }
            }else{
                right --;
                while(left < right && height[right] < mn){
                    res += mn - height[right--];//
                }
            }
        }
        return res;
    }
};