• LeetCode 64 最小路径和


    Leetcode 64 最小路径和

    典型的动态规划问题

    /**动态规划
             * 1. DP[i][j]表示从起点(0,0)到(i,j)位置的最小路径
             * 2. DP[i][j]只与DP[i-1][j]、DP[i][j-1]有关
             * 3. DP[i][j] = Min(DP[i-1][j], DP[i][j-1]) + grid[i][j]
             */
    
    /*第一种: 二维DP数组*/
    //用二维数组存储grid中每一个位置上的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];
    
            /*状态递推: 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];
        }
    }
    
    /*第二种: 一维DP数组*/
    //用一维数组每次存储上一行的DP结果,则下一行DP结果直接在原数组基础上存储
    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];
    
            /*状态递推: 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];
        }
    }
    
    
    /*第三种: 无需额外空间*/
    class Solution {
        public int minPathSum(int[][] grid) {
            if(grid==null || grid.length==0 || grid[0].length==0) return 0;
    
            /*状态递推: 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];
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/CodeSPA/p/13347417.html
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