Leetcode 64 最小路徑和

典型的動態規劃問題

/**動態規劃
         * 1. DP[i][j]表示從起點(0,0)到(i,j)位置的最小路徑
         * 2. DP[i][j]只與DP[i-1][j]、DP[i][j-1]有關
         * 3. DP[i][j] = Min(DP[i-1][j], DP[i][j-1]) + grid[i][j]
         */

/*第一種: 二維DP陣列*/
//用二維陣列儲存grid中每一個位置上的DP值
class Solution {
    public int minPathSum(int[][] grid) {
        if(grid==null || grid.length==0 || grid[0].length==0) return 0;

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

        /*狀態遞推: 按行*/
        for(int i=0; i<grid.length; i++){
            for(int j=0; j<grid[0].length; j++){
                if(i==0 && j==0) dp[i][j] = grid[i][j];
                else if(i==0)
                    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] = Math.min(dp[i][j-1], dp[i-1][j]) + grid[i][j];
            }
        }

        return dp[grid.length-1][grid[0].length-1];
    }
}

/*第二種: 一維DP陣列*/
//用一維陣列每次儲存上一行的DP結果，則下一行DP結果直接在原陣列基礎上儲存
class Solution {
    public int minPathSum(int[][] grid) {
        if(grid==null || grid.length==0 || grid[0].length==0) return 0;

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

        /*狀態遞推: 按行*/
        for(int i=0; i<grid.length; i++){
            for(int j=0; j<grid[0].length; j++){
                if(i==0 && j==0) dp[j] = grid[i][j];
                else if(i==0)
                    dp[j] = dp[j-1] + grid[i][j];
                else if(j==0)
                    dp[j] = dp[j] + grid[i][j];
                else
                    dp[j] = Math.min(dp[j-1], dp[j]) + grid[i][j];
            }
        }

        return dp[grid[0].length-1];
    }
}


/*第三種: 無需額外空間*/
class Solution {
    public int minPathSum(int[][] grid) {
        if(grid==null || grid.length==0 || grid[0].length==0) return 0;

        /*狀態遞推: 按行*/
        for(int i=0; i<grid.length; i++){
            for(int j=0; j<grid[0].length; j++){
                if(i==0 && j==0) continue;
                else if(i==0)
                    grid[i][j] = grid[i][j-1] + grid[i][j];
                else if(j==0)
                    grid[i][j] = grid[i-1][j] + grid[i][j];
                else
                    grid[i][j] = Math.min(grid[i][j-1], grid[i-1][j]) + grid[i][j];
            }
        }

        return grid[grid.length-1][grid[0].length-1];
    }
}