【一天一道LeetCode】#42. Trapping Rain Water

一天一道LeetCode系列

（一）题目

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!

（二）解题

题目需要求的是矩阵作为容器能盛多少体积的水。
方法可以参考我的这篇博客：【一天一道LeetCode】#11Container With Most Water

/*思路：考虑到所盛的水取决于容器两端中小的那一端。因此用两个指针，分别指向头和尾，依次向中间移动。1.如果左边小，则左边向右移动一格，这个时候需要判断向右移动一格后①如果高度大于原来的就表示盛不下水②如果小于原来的则表示有凹下去的部分，这个时候计算高度差就代表能盛多少水。(右边比左边高，可以保证右边不溢出)2.如果右边小，则右边向左移动一格，这个时候同1一样判断。*/class Solution {public:    int trap(vector<int>& height) {        if(height.size()<=2) return 0;        int ret = 0;        int l = 0;        int r = height.size()-1;        int left = height[0];        int right = height[r];        while(l<r)        {            if(left<=right)            {                l++;                if(height[l]>=left)                {                    left = height[l];                }                else ret+=(left-height[l]);            }            else            {                r--;                if(height[r]>=right)                {                    right = height[r];                }                else ret +=(right-height[r]);            }        }        return ret;    }};