Trapping Rain Water--LeetCode

1.题目

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!

2.题意

给定 n 个非负整数，表示每条条形图的宽度是 1，计算出雨后能捕获多少水

3.分析

1）
基于动态规划 Dynamic Programming
维护一个一维数组dp
遍历数组两次，找出i位置左边的最大值和右边最大值
取其较小值，与height[i]相比
如果大于height[i]，则将差值存入结果
注意第二次遍历需要先将maxH重置为0

2）
遍历数组一次即可
先比较两头找出较小值
如果较小值是left，则从左向右扫描
如果较小值是right，则从右向左扫描
若遇到的值比较小值小，则将差值存入结果，
如遇到的值大，则重新确定新的窗口范围，直至left和right指针重合

4.代码

注意res都为+=得到，不要错写成=
1）

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

2）

class Solution {public:    int trap(vector<int>& height) {        int len = height.size();        if(len == 0)            return 0;        int l = 0;        int r = len - 1;        int res = 0;        while(l < r)        {            int minH = min(height[l], height[r]);            if(height[l] == minH)            {                ++l;                while(l < r && height[l] < minH)                {                    res += minH - height[l];                    ++l;                }            }            else            {                --r;                while(l < r && height[r] < minH)                {                    res += minH - height[r];                    --r;                }            }        }        return res;    }};