[leetcode] 42. Trapping Rain Water 解题报告
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题目链接: https://leetcode.com/problems/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!
思路: two point指向左右两端, 维护一个当前左右同时能达到的最大的高度, 并且让面积小的一端先移动, 如果遇到一个新的高度小于当前最高高度, 那么其高度差就可以存储一些水. 如果碰到一个高度, 使得左右能同时达到的高度增加, 则更新.
代码如下:
class Solution {public: int trap(vector<int>& height) { int left = 0, right = height.size()-1, ans =0, Min =0; while(left <= right) { int val = min(height[left], height[right]); if(val < Min) ans += (Min - val); else Min = val; height[left]<=height[right]?left++:right--; } return ans; }};
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