[LeetCode] Triangle,

　　

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

For example, given the following triangle

[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]

 The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).

 

Thoughts:

It's a typical Dynamic Programming question, make sure be able to understand and solve it in mind,

the formula should be: 

 dp[i][j] = dp[i+1][j] + dp[i+1][j+1]

 

it is only correct and easier to solve it from bottom-up, the minimum value for theith value in a certain row equals to the value of the ith index plus the minimum value you can get between the ith and (i+1)th value from the lower row. since it is a Triangle, and the historical data from the lists can only be access once, we don't even need a two-dimensional array, then the formula becomes

dp[j]=triangle.get(i).get(j) + Math.min(dp[j],dp[j+1]);

 

So, the final code should be:

   public int minimumTotal(List<List<Integer>> triangle) {
     int length = triangle.size();
     if(length==0) return 0;
     if(length==1) return triangle.get(0).get(0);
    
     int[] dp = new int [length]; 
     
     //initialize with the values from the last row
     for(int i=0;i<triangle.get(length-1).size();i++){
         dp[i]=triangle.get(length-1).get(i);
      }
     
     //loop from the second last row
     for(int i=length-2;i>=0;i--){
         for(int j=0;j<triangle.get(i).size();j++)
             dp[j]=triangle.get(i).get(j) + Math.min(dp[j],dp[j+1]);
     }
     return dp[0];
    }