[LeetCode] Trapping Rain Water

標籤：leetcode   c++   

Trapping Rain Water

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!

解題思路：

1、我的辦法：兩邊夾的思想，若左邊小於右邊，則移動左邊的指標，否則移動右邊的指標。同時，若移動的指標小於上一個最小高度，則容量減去該移動指標此時所指的高度。若移動的指標和未移動的指標都大於此前最小高度，則加上超出的部分。說的不明白，代碼比較好理解：

class Solution {public:int trap(vector<int>& height) {int len = height.size();if (len<3){return 0;}int start = 0, end = len - 1;int result = 0, minH = 0;minH = min(height[start], height[end]);result = minH * (end - start - 1);while (start<end - 1){if (height[start]<height[end]){start++;result = result - min(minH, height[start]);}else{end--;result = result - min(minH, height[end]);}if (height[start]>minH && height[end]>minH){result += min(height[start] - minH, height[end] - minH) * (end - start - 1); //加上超出的部分minH = min(height[start], height[end]);}}return result;}};

2、網上另外一個比較好的辦法是左右兩邊掃的辦法。記錄每個數左邊和右邊最大的數。左邊最大與右邊最大構成一個容器，減去本身的空間，這相當於這個位置上的台階貢獻了容量。

class Solution {public:int trap(vector<int>& height) {int len = height.size();if (len<3){return 0;}int maxL[len], maxR[len];int maxH = height[0];maxL[0] = 0;for(int i=1; i<len; i++){    maxL[i] = maxH;    maxH = max(maxH, height[i]);}maxH = height[len-1];maxR[len-1]=0;int result=0;for(int i=len-2; i>=0; i--){    maxR[i]=maxH;    maxH = max(maxH, height[i]);        result +=max(0, min(maxL[i], maxR[i]) - height[i]);}return result;}};

[LeetCode] Trapping Rain Water