【LeetCode】42. Trapping Rain Water

Trapping Rain Water

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!

双指针left,right分别从首尾开始扫，记当前left指针遇到的最大值为leftWall，right指针遇到的最大值为rightWall

（1）leftWall <= rightWall

left前进一个位置。

对于left指针指向的位置，若存在被trap，则被trap的值为(leftWall-A[left])。

解释如下：

a.如果left与right之间不存在比leftWall大的值，那么i位置trap的值就取决与leftWall与rightWall的较小值，也就是leftWall

b.如果left与right之间存在比leftWall大的值，其中离leftWall最近的记为newLeftWall，那么i位置trap的值就取决与leftWall与newLeftWall的较小值，也就是leftWall

（2）leftWall > rightWall

right后退一个位置。

对于right指针指向的位置，被trap的值为(rightWall-A[right])。

解释同上。

class Solution {
public:
    int trap(int A[], int n) {
        int ret = 0;
        int left = 0;
        int right = n-1;
        int leftWall = A[left];
        int rightWall = A[right];
        while(left < right)
        {
            if(leftWall <= rightWall)
            {
                left ++;
                if(A[left] <= leftWall)
                    ret += (leftWall - A[left]);
                else
                    leftWall = A[left];
            }
            else
            {
                right --;
                if(A[right] <= rightWall)
                    ret += (rightWall - A[right]);
                else
                    rightWall = A[right];
            }
        }
        return ret;
    }
};