Leetcode题解42-Trapping Rain Water

题目：

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.

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 Marcosfor contributing this image!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

思路：双指针从左右遍历，维护当前位置所对应的左侧最高值和右侧最高值，这两者中的最小值与当前位置的差值即为可储水量

代码

class Solution {
public:
    int trap(vector<int>& height) {
        int Len = height.size();
        if(Len <= 2) return 0;   //需要三个以上才能储水
        vector<int> leftArr(Len,0), rightArr(Len,0);   // 维护两侧最高值的数组
        int maxLeft=0, maxRight=0, sumWater = 0;
        for(int i = 0; i <height.size(); i++){
            leftArr[i] = max(height[i],maxLeft);     
            maxLeft = max(height[i],maxLeft);   //更新MaxLeft
        }
        for(int i = Len-1; i >=0; i--){
            rightArr[i] = max(height[i], maxRight);
            maxRight = max(height[i], maxRight);
        }     
        for(int i = 0;  i < Len; i++){
            sumWater += min(leftArr[i], rightArr[i]) - height[i];
        }
        return sumWater;        
    }
};