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Trapping Rain Water

 

Total Accepted: 14568 

Total Submissions: 50810

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!

题意：有一个代表 n 根柱子高度的数组A[i]，计算这些柱子之间的槽可以储蓄的最大水量

思路：左右 dp

用一个数组 left[i] 表示第 i 根柱子左边最高的柱子的高度

用一个数组 right[i] 表示第 i 根柱子右边最高的柱子的高度

1.从左到右扫描。left[i] = max(left[i - 1], A[i - 1])

2.从右到左扫描。right[i] = max(right[i + 1], A[i + 1])

3.第 i 根柱子上能储蓄的水为 min(left[i], right[i]) - A[i]，由于这时 left[i] 已经没用了，能够用它来存放这个值。

复杂度：时间O(n)。 空间O(n)

相关题目：

Candy

int trap(int A[], int n){	if (n == 3) return 0;	vector
     
       left(n, 0), right(n, 0);	for(int i = 1; i < n - 1; ++i) left[i] = max(left[i - 1], A[i - 1]);	for(int i = n - 2; i > 0; --i) {		right[i] = max(right[i + 1], A[i + 1]);		left[i] = min(left[i], right[i]) - A[i];	}	int sum = 0;	for_each(left.begin() + 1, left.end() - 1, [&sum](int c){		if(c > 0) sum += c;	});	return sum ;}