LintCode:最小路径和

最小路径和

给定一个只含非负整数的m*n网格，找到一条从左上角到右下角的可以使数字和最小的路径。

LintCode:最小路径和

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思路一：万能的枚举

• m行，n列的矩阵，从左上角走到右下角，需要向下移动(m-1)步，向右移动(n-1)步，共（m+n-2）步，则路径总条数为C(m+n-2， m-1)步。

• m,n较小时可行，较大时不可行。

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思路二： 动态规划

• dp[i][j] 表示从左上到达grid[i][j]的最小值。
• dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + a[i][j]
  
  • 从上面过来为dp[i-1][j] + grid[i][j]
  • 从左边过来为dp[i][j-1] + grid[i][j]
• 初值
  
  • dp[0][0] = grid[0][0]
  • dp[0][j>0] = dp[0][j-1] + grid[0][j]
  • dp[i>0][0] = dp[i-1][0] + grid[i][0]
• 复杂度：时间O(m*n), 空间O(m*n)

class Solution {public:    /**     * @param grid: a list of lists of integers.     * @return: An integer, minimizes the sum of all numbers along its path     */    int minPathSum(vector<vector<int> > &grid) {        // write your code here        int m = grid.size();        int n = grid[0].size();        vector<vector<int> > dp(m, vector<int>(n));        for (int i=0; i<m; i++){            for (int j=0; j<n; j++){                if (i==0){                    if(j==0){                        dp[i][j] = grid[i][j];                    }                    else{                        dp[i][j] = dp[i][j-1] + grid[i][j];                    }                }                else if(j==0){                    dp[i][j] = dp[i-1][j] + grid[i][j];                }                else{                    dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j];                }            }        }        return dp[m-1][n-1];    }};

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dp空间优化

• dp[i][j]只与dp[i-1][j], dp[i][j-1]有关

• 对每个i， 正向循环j

  • 前的dp[j-1]是新的，dp[j]还是旧的
  • dp[j] = min(dp[j-1], dp[j] + a[i][j] 更新

• 空间复杂度O(n), 时间复杂度O(m*n)

class Solution {public:    /**     * @param grid: a list of lists of integers.     * @return: An integer, minimizes the sum of all numbers along its path     */    int minPathSum(vector<vector<int> > &grid) {        // write your code here        int m = grid.size();        int n = grid[0].size();        vector<int > dp(n);        for (int i=0; i<m; i++){            for (int j=0; j<n; j++){                if (i==0){                    if(j==0){                        dp[j] = grid[i][j];                    }                    else{                        dp[j] = dp[j-1] + grid[i][j];                    }                }                else if(j==0){                    dp[j] = dp[j] + grid[i][j];                }                else{                    dp[j] = min(dp[j-1], dp[j]) + grid[i][j];                }            }        }        return dp[n-1];    }};

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参考资料

七月算法公开课 实战动态规划