動(dòng)態(tài)規(guī)劃的解法

class Solution {
    public int minPathSum(int[][] grid) {
        if (grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        // 設(shè)f[i][j]為從左上角到坐標(biāo)(i, j)的最小路徑和
        // 狀態(tài)轉(zhuǎn)移方程: f[i][j] = min{ f[i - 1][j], f[i][j - 1] } + grid[i][j]
        // 初始條件:    f[0][0] = grid[0][0]
        // 邊界條件:
        // 1. f[0][j] = f[0][j - 1] + grid[i][j]
        // 2. f[i][0] = f[i - 1][0] + grid[i][j]
        int m = grid.length;
        int n = grid[0].length;
        int f[][] = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0 && j == 0) {
                    f[i][j] = grid[0][0];
                } else if (i == 0) {
                    f[i][j] = f[i][j - 1] + grid[i][j];
                } else if (j == 0) {
                    f[i][j] = f[i - 1][j] + grid[i][j];
                } else {
                    f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + grid[i][j];
                }
            }
        }
        return f[m - 1][n - 1];
    }
}

空間壓縮

由于第i行的f[i][j]只由第i行和第i - 1行的數(shù)據(jù)所決定，所以我們可以進(jìn)行空間復(fù)用，把空間復(fù)雜度從O(m * n)壓縮到O(n)（其中m是行數(shù)，n是列數(shù)）。在動(dòng)態(tài)規(guī)劃中，壓縮行或是列取決于哪個(gè)更大。

class Solution {
    public int minPathSum(int[][] grid) {
        if (grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        // 設(shè)f[i][j]為從左上角到坐標(biāo)(i, j)的最小路徑和
        // f[i][j] = min{ f[i - 1][j], f[i][j - 1] } + a[i][j]
        int m = grid.length;
        int n = grid[0].length;
        // 只有兩個(gè)狀態(tài)，所以只開兩行的空間
        int f[][] = new int[2][n];
        // 設(shè)兩個(gè)滾動(dòng)的狀態(tài)
        // old: i - 1
        // now: i
        int old, now = 0;
        for (int i = 0; i < m; i++) {
            // 對(duì)于每一個(gè)新行，交換old和now
            old = now;
            now = 1 - now;
            for (int j = 0; j < n; j++) {
                if (i == 0 && j == 0) {
                    f[now][j] = grid[0][0];
                } else if (i == 0) {
                    f[now][j] = f[now][j - 1] + grid[i][j];
                } else if (j == 0) {
                    f[now][j] = f[old][j] + grid[i][j];
                } else {
                    f[now][j] = Math.min(f[old][j], f[now][j - 1]) + grid[i][j];
                }
            }
        }
        return f[now][n - 1];
    }
}

如何打印路徑

class Solution {
    public int minPathSum(int[][] grid) {
        if (grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        final int m = grid.length;
        final int n = grid[0].length;
        int[][] f = new int[m][n];

        // decision[i][j] = 'U'：f(i, j)使用的是來(lái)自上方的sum
        // decision[i][j] = 'L'：f(i, j)使用的是來(lái)自左方的sum
        char[][] decision = new char[m][n];

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0 && j == 0) {
                    f[i][j] = grid[0][0];
                } else if (i == 0) {
                    f[i][j] = f[i][j - 1] + grid[i][j];
                    decision[i][j] = 'L';
                } else if (j == 0) {
                    f[i][j] = f[i - 1][j] + grid[i][j];
                    decision[i][j] = 'U';
                } else {
                    int t = Math.min(f[i - 1][j], f[i][j - 1]);
                    f[i][j] = t + grid[i][j];
                    decision[i][j] = t == f[i - 1][j] ? 'U' : 'L';
                }
            }
        }
        // 從左上走到右下有多少步？m + n - 1步
        int[] path = new int[m + n - 1];
        int i = m - 1;
        int j = n - 1;
        // 根據(jù)決策從最后一步還原到第一步
        for (int p = m + n - 2; p >= 0; p--) {
            path[p] = grid[i][j];
            if (decision[i][j] == 'U') {
                i--;
            } else if (decision[i][j] == 'L') {
                j--;
            }
        }
        // 從第一步打印到最后一步
        for (int p = 0; p < m + n - 1; p++) {
            System.out.print(path[p] + " ");
        }
        System.out.println();

        return f[m - 1][n - 1];
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        int[][] grid = {{1, 5, 7, 6, 8}, {4, 7, 4, 4, 9}, {10, 3, 2, 3, 2}};
        System.out.println(solution.minPathSum(grid));
    }
}