问题描述
Trapping Rain Water
Givennnon-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], return6.
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 Marcosfor contributing this image!
问题分析
这道题目解题思路可以参考leetcode011：Container With Most Water的分析，leetcode011是求围住的最大面积，这里有所区别，，统计“装水”面积，还是从两边向中间扫描，时间复杂度O(n)。 这里设置一个水平面h，表示当前围住的高度，如果后续扫面到的比h低，说明可以装水，统计；如果比h高，更新h即可。
代码
//运行时间：13msclass Solution {public:int trap(vector<int>& height) {int i = 0, j = height.size()-1;int ans = 0;int h = 0;while (j-i >= 0){if (height[i] > height[j]){if (height[j] <= h){ans += (h – height[j]);}else{h = height[j];}j–;}else if (height[i] < height[j]){if (height[i] <= h){ans += (h – height[i]);i++;}else{h = height[i];}}else{if (height[i] <= h){if (i != j)ans += 2 * (h – height[i]);elseans += (h – height[i]);}else{h = height[i];}i++; j–;}}return ans;}};
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