lintcode-110-最小路径和

110-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。

注意事项

你在同一时间只能向下或者向右移动一步

样例

标签

动态规划

思路

使用动态规划，用二维数组 dp[i][j] 表示网格第 i 行、第 j 列元素到右下角的最小路径

动态转移方程为：dp[i][j] = gird[i][j] + min(dp[i+1][j], dp[i][j+1])

过程如下图

code

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 sizeRow = grid.size(), sizeCol = grid[0].size(), i = 0, j = 0;
        if(sizeRow <= 0) {
            return 0;
        }

        vector<vector<int> > dp(sizeRow, vector<int>(sizeCol, 0x7FFFFFFF));

        for(i=sizeRow-1; i>=0; i--) {
            for(j=sizeCol-1; j>=0; j--) {
                if(i == sizeRow-1 && j == sizeCol-1) {
                    dp[i][j] = grid[i][j];
                }
                else if(i == sizeRow-1 && j < sizeCol-1) {
                    dp[i][j] = grid[i][j] + dp[i][j+1];
                }
                else if(i < sizeRow-1 && j == sizeCol-1) {
                    dp[i][j] = grid[i][j] + dp[i+1][j];
                }
                else if(i < sizeRow-1 && j < sizeCol-1) {
                    dp[i][j] = grid[i][j] + ((dp[i+1][j] > dp[i][j+1]) ? dp[i][j+1] : dp[i+1][j]);
                }
            }
        }

        return dp[0][0];
    }
};