[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]

, 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!

在上述结构中，在两片栅栏间 装水的多少取决于

1.独立的容器，那么取决于最短的那块木板。

2.非独立的容器，即被包在更大的容器中，那么装水的多少取决于更大的容器。

按照以上思路 我们只有从两边向中间靠拢，才可以获得最优解法。

从两边往中间，每次选取最短的木板向中间遍历，就能保证每片低于边界的地方 水都是可以被保存的。

class Solution {
public:
int trap(int A[], int n) {
int i = 0, j = n - 1, res = 0, cur;
int *a = A;
while (i < j) {
if (a[i] < a[j]) {
cur = i+1;
while (cur <= j && a[cur] <= a[i]) res += a[i] - a[cur++];
i = cur;
} else {
cur = j - 1;
while (cur >= i && a[cur] <= a[j]) res += a[j] - a[cur--];
j = cur;
}
}
return res;
}
};