leetcode problem 42 -- 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!

 

 

思路:

　　这一道题类似于leetcode 11 题Container With Most Water.  要求最多能装多少水.有两种思路.

　　思路一:

　　　　先从左往右遍历.每遍历一个数字,就贪心得到这个数字所在的格子能最多装水. 由于是从左到右遍历,我们只考虑的左边的高度,没有考虑右边的高度(实际应该选择两者小的那个). 因此需要再从右往左遍历,同样贪心这个数字所在格子

　　最多能装多少水,这次以右边高度为准. 最终综合从左到右和从右到左的结果,取两者小的.

 

代码:(runtime 19ms)

 1 class Solution {

 2 public:

 3     int trap(vector<int>& height) {

 4         vector<int> leftToRight;

 5         vector<int> rightToLeft;

 6         aux_function(height.begin(), height.end(), leftToRight);

 7         

 8         aux_function(height.rbegin(), height.rend(), rightToLeft);

 9 

10         int ans = 0;

11         auto it1 = leftToRight.begin();

12         auto it2 = rightToLeft.rbegin();

13 

14         while (it1 != leftToRight.end()){

15             ans += min(*it1++, *it2++);

16         }

17         return ans;

18     }

19 

20 private:

21 template<class Iter>

22     void aux_function(Iter begin, Iter end, vector<int> &ret) {

23         int left = 0;

24         for (auto it = begin; it != end; it++) {

25             left = left > *it ? left : *it;

26             ret.push_back(left - *it);

27         }

28     }

29 };

View Code

 

　　思路二:

　　　　如上图中所示, 先求黑色格子与蓝色格子的总面积,然后再减去黑色各自面积.

代码:(runtime 10ms)

　　

class Solution {

public:

    int trap(vector<int> A) {

        int n = A.size();

        int summap = 0;

        int sumtot = 0;



        for(int i = 0; i < n; i++) summap += A[i];



        int left = 0, right = n - 1;

        int leftbar = 0, rightbar = 0;

        while(left <= right) {

            leftbar = max(A[left], leftbar);

            rightbar = max(A[right], rightbar);



            if(leftbar <= rightbar) {

                sumtot += leftbar;

                left++;

                //right--;

            } else {

                sumtot += rightbar;

                right--;

                //left++;

            }

        }



        return sumtot - summap;

    }

};

View Code