[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。

class Solution {
public:
    int trap(vector<int>& height) {
        int length = height.size();
        vector<int> max_left(length, 0);
        vector<int> max_right(length, 0);
        
        int result = 0;
        
        for (int i = 1; i < length; ++i) {
            max_left[i] = max(max_left[i - 1], height[i - 1]);
            max_right[length - 1 - i] = max(max_right[length - i], height[length - i]);
        }
        
        for (int i = 0; i < length; ++i) {
            int h = min(max_left[i], max_right[i]);
            if (h > height[i]) {
                result += (h - height[i]);
            }
        }
        
        return result;
    }
};