Sword finger offer II 091 Paint the house

Looked at the leetcode Today's A daily topic , Topic link ： The finger of the sword Offer II 091. Paint the house
The title is as follows ：

If there is a row of houses , common n individual , Every house can be painted red 、 One of the three colors blue or green , You need to paint all the houses and make them adjacent The two house colors Cannot be the same .

Of course , Because the prices of different colors of paint on the market are different , So the cost of painting the house in different colors is different . The cost of painting each house in a different color is one n x 3 Positive integer matrix of costs To represent the .

for example ,costs[0][0] It means the first one 0 The cost of painting the house red ;costs[1][2] It means the first one 1 The cost of painting the house green , And so on .

Please figure out that all the houses have been painted least The cost of .

I use the dynamic programming method here , The equation is as follows ：
1. initialization dp Array

dp[0][0] = costs[0][0];
dp[0][1] = costs[0][1];
dp[0][2] = costs[0][2];

2. State transition equation ：i The range of [1,n)

dp[i][0] = Math.min(dp[i-1][1],dp[i-1][2])+costs[i][0];
dp[i][1] = Math.min(dp[i-1][0],dp[i-1][2])+costs[i][1];
dp[i][2] =Math.min(dp[i-1][0],dp[i-1][1])+costs[i][2];

The overall code is as follows ：

    public int minCost(int[][] costs) {
    

        int n = costs.length;
        int len = costs[0].length;

        int[][]  dp = new int[n][len];

        dp[0][0] = costs[0][0];
        dp[0][1] = costs[0][1];
        dp[0][2] = costs[0][2];

        for(int i = 1;i < n;i++){
    
           
            dp[i][0] = Math.min(dp[i-1][1],dp[i-1][2])+costs[i][0];
            dp[i][1] = Math.min(dp[i-1][0],dp[i-1][2])+costs[i][1];
            dp[i][2] = Math.min(dp[i-1][0],dp[i-1][1])+costs[i][2];
            
        }
        return Math.min(dp[n-1][0],Math.min(dp[n-1][1],dp[n-1][2]));

    }

Of course, you can also change the ternary operator to try .
Minor code changes , Try to save some space , The code is as follows ：

    public int minCost(int[][] costs) {
    

        int n = costs.length;
        int len = costs[0].length;

        int[][]  dp = new int[2][len];

        dp[0][0] = costs[0][0];
        dp[0][1] = costs[0][1];
        dp[0][2] = costs[0][2];

        for(int i = 1;i < n;i++){
    
           
            dp[1][0] = Math.min(dp[0][1],dp[0][2])+costs[i][0];
            dp[1][1] = Math.min(dp[0][0],dp[0][2])+costs[i][1];
            dp[1][2] = Math.min(dp[0][0],dp[0][1])+costs[i][2];
            dp[0][0] = dp[1][0];
            dp[0][1] = dp[1][1];
            dp[0][2] = dp[1][2];

            dp[1][0] = 0;
            dp[1][0] = 0;
            dp[1][0] = 0;
            
        }
        return Math.min(dp[0][0],Math.min(dp[0][1],dp[0][2]));

    }