• 【leetcode】Triangle


    Question:

    Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

    For example, given the following triangle

    [
         [2],
        [3,4],
       [6,5,7],
      [4,1,8,3]
    ]

    Anwser 1:      

    class Solution {
    public:
        int minimumTotal(vector<vector<int> > &triangle) {
            // Start typing your C/C++ solution below
            // DO NOT write int main() function
            int rows = triangle.size();
            if(0 == rows) return 0;
            
            int *minSums = new int[rows];
            int *temp = new int[rows];
            
            for(int r = 0; r < rows; r++) {
                vector<int> vec = triangle[r];
                temp[0] = vec[0] + (r > 0 ? minSums[0] : 0);
                for(int i = 1; i < r; i++) {
                    temp[i] = vec[i] + min(minSums[i-1], minSums[i]);
                }
                
                if(r > 0) {
                    temp[r] = vec[r] + minSums[r-1];
                }
                    
                int *tswap = temp;
                temp = minSums;
                minSums = tswap;
            }
            
            int m = minSums[0];
            for(int i = 1; i < rows; i++) 
            {
                if(minSums[i] < m){
                    m = minSums[i];
                } 
            }
            
            delete temp;
            delete minSums;
            return m;
        }
    };


    Anwser 2:     

    class Solution {
    public:
        int minimumTotal(vector<vector<int> > &triangle) {
            // Start typing your C/C++ solution below
            // DO NOT write int main() function
            int line = triangle.size();
    
            for(int i = line -2 ; i >= 0; i--) 
            {
                for(int j = 0; j < triangle[i].size(); j++)
                {
                    triangle[i][j] += min(triangle[i+1][j], triangle[i+1][j+1]);
                }
            }
    
            return triangle[0][0];
        }
    };


    Anwser 3:

    class Solution {
    public:
        
        void run(vector<vector<int> > &triangle, int row, int idx, int curSum, int &minPath)
        {
            if (row == triangle.size())
            {
                minPath = min(minPath, curSum);
                return;
            }
            
            run(triangle, row + 1, idx, curSum + triangle[row][idx], minPath);
            run(triangle, row + 1, idx + 1, curSum + triangle[row][idx], minPath);
        }
    
        int minimumTotal(vector<vector<int> > &triangle) {
            // Start typing your C/C++ solution below
            // DO NOT write int main() function
            int minPath = INT_MAX;
            
            if (triangle.size() == 0) {
                return 0;
            }
            
            run(triangle, 0, 0, 0, minPath);
            
            return minPath;
        }
    };

    注意点:

    1) Judge Small is ok, but Judge Large is error

    2) 如果迭代很深的话,容易造成压栈占用内存很高,超出段后会溢出

  • 相关阅读:
    ubuntu下安装JDK(复制)
    idea的ssm搭建(复制)
    linux常用命令(复制)
    Ubuntu安装nginx(复制)
    win7分盘(复制)
    mysql环境变量配置(复制)
    mysql的下载及配置(复制1)
    java环境变量的配置
    Windows 右键添加「cmd 打开」
    快速开启Windows 的各种任务及 bat(ch)脚本
  • 原文地址:https://www.cnblogs.com/xinyuyuanm/p/3020136.html
Copyright © 2020-2023  润新知