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  • 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
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  • 原文地址:https://www.cnblogs.com/lysuns/p/4481918.html
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