【LeetCode】Trapping Rain Water

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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!

 

雙指標i,j分別從首尾開始掃，記當前i指標遇到的最大值為leftWall，j指標遇到的最大值為rightWall

（1）leftWall <= rightWall

對於i指標指向的位置，被trap的值為(leftWall-A[i])。

i前進一個位置。

解釋如下：

a.如果i與j之間不存在比leftWall大的值，那麼i位置trap的值就取決與leftWall與rightWall的較小值，也就是leftWall

b.如果i與j之間存在比leftWall大的值，其中離leftWall最近的記為newLeftWall，那麼i位置trap的值就取決與leftWall與newLeftWall的較小值，也就是leftWall

（2）leftWall > rightWall

對於j指標指向的位置，被trap的值為(rightWall-A[j])。

j前進一個位置。

解釋同上。

 

class Solution {public:    int trap(int A[], int n) {        if(n < 3)   //at least 3 integers            return 0;                int i = 0;        int j = n-1;        int leftWall = 0;        int rightWall = 0;        int ret = 0;        while(i <= j)        {            //update leftWall and rightWall            leftWall = max(leftWall, A[i]);            rightWall = max(rightWall, A[j]);            if(leftWall <= rightWall)            {//no matter whether there is a more higher wall between leftWall and rightWall             //A[i] is dependent on leftWall if leftWall <= rightWall                ret += (leftWall-A[i]);                i ++;            }            else            {//no matter whether there is a more higher wall between leftWall and rightWall             //A[j] is dependent on rightWall if rightWall < leftWall                ret += (rightWall-A[j]);                j --;            }        }        return ret;    }};

【LeetCode】Trapping Rain Water