題目

https://leetcode-cn.com/problems/minimum-path-sum/
給定一個(gè)包含非負(fù)整數(shù)的 m x n 網(wǎng)格，請(qǐng)找出一條從左上角到右下角的路徑吏垮，使得路徑上的數(shù)字總和為最小障涯。

說明：每次只能向下或者向右移動(dòng)一步。
難度：中等

思路

1.動(dòng)態(tài)規(guī)劃膳汪，開辟一個(gè)mxn的數(shù)組唯蝶，存儲(chǔ)當(dāng)前的最小值，最小值實(shí)際就是左和上兩者中小較小的加上當(dāng)前的值遗嗽。
2.由于每個(gè)用過的元素不會(huì)二次使用粘我，因此可以直接更新當(dāng)前的數(shù)組，不用新開辟痹换，可以節(jié)省內(nèi)存

解法

1.動(dòng)態(tài)規(guī)劃征字。

//leetcode submit region begin(Prohibit modification and deletion)
class minPathSumx {
    public int minPathSum(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }

        int rows = grid.length, columns = grid[0].length;
        int[][] dp = new int[rows][columns];
        dp[0][0] = grid[0][0];
        for (int i = 1; i < rows; i++) {
            dp[i][0] = dp[i - 1][0] + grid[i][0];
        }
        for (int j = 1; j < columns; j++) {
            dp[0][j] = dp[0][j - 1] + grid[0][j];
        }
        for (int i = 1; i < rows; i++) {
            for (int j = 1; j < columns; j++) {
                dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
            }
        }
        return dp[rows - 1][columns - 1];
}
//leetcode submit region end(Prohibit modification and deletion)

2.動(dòng)態(tài)規(guī)劃都弹，不開辟dp。

//leetcode submit region begin(Prohibit modification and deletion)
class minPathSumx {
    public int minPathSum(int[][] grid) {
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < columns; j++) {
                if (0 == i && 0 == j) continue;
                else if (0 == i) grid[i][j] += grid[i][j - 1];
                else if (0 == j) grid[i][j] += grid[i - 1][j];
                else grid[i][j] += Math.min(grid[i - 1][j], grid[i][j - 1]);
            }
        }
        return grid[rows - 1][columns - 1];
}
//leetcode submit region end(Prohibit modification and deletion)