[leetcode] 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], 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
;
int[] rightMax = new int
;

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