【Leetcode】Traping Rain Water (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!

首先定义一个container长度和A相同，container用于装水。A数组用于放石头

怎么测试装水呢，先找container中每个点container[i]的瓶颈高度

首先从左往右找每一个点的左边最大高度 0 0 1 1 2 2 2 2 3 3 3 3

然后从右往左找每一个点的右边最大高度 3 3 3 3 3 3 3 2 2 2 1 0

最后根据左右最大值找每个点的瓶颈高度 0 0 1 1 2 2 2 2 2 2 1 0

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最后是每一个被大石头占领了的高度A[ ]   0 1 0 2 1 0 1 3 2 1 2 1

所以 只要container的瓶颈高度container[i]大于A[i]的高度，就说明i这里是可以装水的。

全部加起来就可以了。

public int trap(int[] A) {if (A == null || A.length == 0)return 0;int[] container = new int[A.length];int max = 0;int result = 0;for (int i = 0; i < A.length; i++) {container[i] = max;max = Math.max(max, A[i]);}max = 0;for (int i = A.length - 1; i >= 0; i--) {container[i] = Math.min(container[i], max);max = Math.max(max, A[i]);if (container[i] > A[i])result += container[i] - A[i];}return result;}