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], 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 Marcos for contributing this image!

https://oj.leetcode.com/problems/trapping-rain-water/

思 路1：对某个值A[i]来说，能装的最多的水取决于在i之前最高的值leftMaxt[i]和在i右边的最高的rightMax[i]，容水量即 min(left,right) – A[i]。:为了计算高度，第一遍从左到右计算数组leftMaxt，第二遍从右到左计算rightMax。时空复杂度都是O(n)。

思路2：改进思路1，可以用常量空间搞定。具体思路，先找到最高的bar，然后总两边向中间计算，边计算边更新curHeight。

具体可参考 这里

思路1：

public class Solution {
	public int trap(int[] A) {
		if (A == null || A.length < 3)
			return 0;
		int n = A.length;
		int[] leftMax = new int[n];
		int[] rightMax = new int[n];

		int i;
		int max = A[0];
		for (i = 1; i < n - 1; i++) {
			leftMax[i] = max;
			if (A[i] > max)
				max = A[i];

		}

		max = A[n - 1];
		for (i = n - 2; i > 0; i--) {
			rightMax[i] = max;
			if (A[i] > max)
				max = A[i];

		}

		int count = 0;

		for (i = 1; i < n - 1; i++) {
			int min = leftMax[i] < rightMax[i] ? leftMax[i] : rightMax[i];
			if (min > A[i])
				count += min - A[i];

		}
		return count;
	}

	public static void main(String[] args) {
		System.out.println(new Solution().trap(new int[] { 0, 1, 0, 2, 1, 0, 1,
				3, 2, 1, 2, 1 }));

	}

}

参考：

http://fisherlei.blogspot.com/2013/01/leetcode-trapping-rain-water.html

http://www.cnblogs.com/lichen782/p/Leetcode_Trapping_Rain_Water.html

[leetcode] Trapping Rain Water,布布扣,bubuko.com

[leetcode] Trapping Rain Water