Trapping Rain Water
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一. 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.
Difficulty:Hard
TIME:20MIN
解法
记得之前做过一道题Container With Most Water,如果有那道题的经验,那么这道题也不难解决。
既然知道积累雨水是基于比较小的边,那么我们从比较小的边开始遍历,直到找到另一个长度大于等于它的边,那么就可以着手计算这个区域内的雨水了(其实采用累加计算,找到长度大于等于它的边,就已经计算了好了这个区域内的雨水)
int trap(vector<int>& height) { if(height.size() < 3) return 0; int result = 0; int left = 0,right = height.size() - 1; int maxLeft = height[left], maxRight = height[right]; while(left <= right) { if(maxLeft >= maxRight) { //从边长比较小的边开始遍历(保证会遇到边长比较大的边) if(height[right] >= maxRight) maxRight = height[right]; //重置边长 else result += maxRight - height[right]; //累加雨水量 right--; } else { if(height[left] >= maxLeft) maxLeft = height[left]; else result += maxLeft - height[left]; left++; } } return result;}
代码的时间复杂度为
0 0
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