[LeetCode] 64. Minimum Path Sum 最小路径和

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

Example:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

给定一个m x n 的含有非负整数的方格，从左上角移动到右下角，每次只能向下或者向右移动，找出最小的路径和。

解法：动态归化(Dynamic Programming)，

State: dp[i][j]，表示从(0, 0)到(i, j)最小的路径和

Function: dp[i][j] = min(dp[i][j-1], dp[i-1][j]) + grid[i][j]

Initialize: dp[0][0] = grid[0][0]

Return: dp[m - 1][n - 1] 

Java:

public class Solution {
    public int minPathSum(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }

        int M = grid.length;
        int N = grid[0].length;
        int[][] dp = new int[M][N];

        dp[0][0] = grid[0][0];

        for (int i = 1; i < M; i++) {
            dp[i][0] = dp[i - 1][0] + grid[i][0];
        }

        for (int i = 1; i < N; i++) {
            dp[0][i] = dp[0][i - 1] + grid[0][i];
        }

        for (int i = 1; i < M; i++) {
            for (int j = 1; j < N; j++) {
                dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
            }
        }

        return dp[M - 1][N - 1];
    }
}

Python:

class Solution:
    # @param grid, a list of lists of integers
    # @return an integer
    def minPathSum(self, grid):
        m = len(grid); n = len(grid[0])
        dp = [[0 for i in range(n)] for j in range(m)]
        dp[0][0] = grid[0][0]
        for i in range(1, n):
            dp[0][i] = dp[0][i-1] + grid[0][i]
        for i in range(1, m):
            dp[i][0] = dp[i-1][0] + grid[i][0]
        for i in range(1, m):
            for j in range(1, n):
                dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]
        return dp[m-1][n-1]　　

Python:

class Solution:
    # @param grid, a list of lists of integers
    # @return an integer
    def minPathSum(self, grid):
        sum = list(grid[0])
        for j in xrange(1, len(grid[0])):
            sum[j] = sum[j - 1] + grid[0][j]

        for i in xrange(1, len(grid)):
            sum[0] += grid[i][0]
            for j in xrange(1, len(grid[0])):
                sum[j] = min(sum[j - 1], sum[j]) + grid[i][j]
                
        return sum[-1]

Python: wo

class Solution(object):
    def minPathSum(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        m, n = len(grid), len(grid[0])
        dp = [[0] * n for i in xrange(m)]
        for i in xrange(m):
            for j in xrange(n):
                if i == 0 and j == 0:
                    dp[i][j] = grid[i][j]
                elif i == 0:
                    dp[i][j] = grid[i][j] + dp[i][j-1]
                elif j == 0:
                    dp[i][j] = grid[i][j] + dp[i-1][j]
                else:
                    dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]
                    
        return dp[-1][-1]     　　

Python:

class Solution:
    """
    @param grid: a list of lists of integers.
    @return: An integer, minimizes the sum of all numbers along its path
    """
    def minPathSum(self, grid):
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if i == 0 and j > 0:
                    grid[i][j] += grid[i][j-1]
                elif j == 0 and i > 0:
                    grid[i][j] += grid[i-1][j]
                elif i > 0 and j > 0:
                    grid[i][j] += min(grid[i-1][j], grid[i][j-1])
        return grid[len(grid) - 1][len(grid[0]) - 1]　　

C++:

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 f[1000][1000];
        if (grid.size() == 0 || grid[0].size() == 0)
            return 0;
        f[0][0] = grid[0][0];
        for(int i = 1; i < grid.size(); i++)
            f[i][0] = f[i-1][0] + grid[i][0];
        for(int i = 1; i < grid[0].size(); i++)
            f[0][i] = f[0][i-1] + grid[0][i];
        for(int i = 1; i < grid.size(); i++)
            for(int j = 1; j < grid[0].size(); j++)
                f[i][j] = min(f[i-1][j], f[i][j-1]) + grid[i][j];
                
        return f[grid.size()-1][grid[0].size()-1];
    }
};

　

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转载于:https://www.cnblogs.com/lightwindy/p/8469442.html