leetcode 120 Triangle ----- java

Posted 2020-08-17 xiaoba1203

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).

Note:
Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

 

 求出最小的路径。

第一次做使用递归，超时了。

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {int len = triangle.size();


        int result = triangle.get(0).get(0);


        result = getResult(result,0,0,triangle);

        
        return result;
        
        
    }

    public static int getResult(int result,int pos,int num,List<List<Integer>> triangle){

        if( num == triangle.size()-1 )
            return result;
        int num1 = triangle.get(num+1).get(pos);
        int ans = result;
        ans += num1;
        ans = getResult(ans,pos,num+1,triangle);


        num1 = triangle.get(num+1).get(pos+1);
        result += num1;
        result = getResult(result,pos+1,num+1,triangle);
        return ans>result?result:ans;

    }
}

 

 

所以还是需要用DP。

 

 比较简单的DP应用。

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        

        int height = triangle.size();

        int[] dp = new int[height];
        dp[0] = dp[0]+triangle.get(0).get(0);
        for( int i = 1;i<height;i++){
            int a = dp[0],b = dp[1];
            dp[0] = dp[0]+triangle.get(i).get(0);
            for( int j = 1;j<i;j++){
                dp[j] = Math.min(a,b)+triangle.get(i).get(j);
                a = b;
                b = dp[j+1];
            }
            dp[i] = a+triangle.get(i).get(i);
        }
        int result = dp[0];
        for( int i = 1;i<height;i++)
            result = Math.min(result,dp[i]);
        
        return result;

    

    }
}