[LeetCode]Trapping Rain Water

题目描述:(链接)

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.

题目解析:

遍历数组，找到当前元素左边的最大值max_left和右边的最大值max_right，如果min(max_left, max_right) < 当前元素高度h，将h - min(max_left, max_right)加到结果上去。

 1 class Solution {
 2 public:
 3     int trap(vector<int>& height) {
 4         int length = height.size();
 5         vector<int> max_left(length, 0);
 6         vector<int> max_right(length, 0);
 7
 8         int result = 0;
 9
10         for (int i = 1; i < length; ++i) {
11             max_left[i] = max(max_left[i - 1], height[i - 1]);
12             max_right[length - 1 - i] = max(max_right[length - i], height[length - i]);
13         }
14
15         for (int i = 0; i < length; ++i) {
16             int h = min(max_left[i], max_right[i]);
17             if (h > height[i]) {
18                 result += (h - height[i]);
19             }
20         }
21
22         return result;
23     }
24 };

转载于:https://www.cnblogs.com/skycore/p/4868909.html

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