Leetcode 动态规划 Trapping Rain Water

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Trapping Rain Water

 Total Accepted: 14568 Total Submissions: 50810My Submissions

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<int> 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 ;
}