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).

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.

 

tag: 经典动态规划。

 

自顶向下， 空间复杂度 O(N^2)

顶底向上，空间复杂度O(N).

 

自顶向下

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        if(triangle == null || triangle.size() == 0) {
            return Integer.MAX_VALUE;
        }
        
        int size = triangle.size();
        
        int[][] f = new int[size][size];
        
        f[0][0] = triangle.get(0).get(0);
        //initialize the diagonal
        for(int i = 1; i < size; i++ ) {
            f[i][i] = f[i - 1][i - 1] + triangle.get(i).get(i);
        }
        
        for(int i = 1 ; i < size; i++) {
            f[i][0] = f[i - 1][0] + triangle.get(i).get(0);
        }
        
        for(int i = 1; i < size; i++){
            for(int j = 1; j < i;j++) {
                f[i][j] = triangle.get(i).get(j) + Math.min(f[i - 1][j - 1], f[i - 1][j]);
            }
        }
        
        int min = f[size - 1][0];
        for(int k = 1; k < size; k++) {
            min = Math.min(min, f[size - 1][k]);
        }
        return min;
        
    }
}

　　

自底向上

public int minimumTotal(List<List<Integer>> triangle) {
    int[] A = new int[triangle.size()+1];
    for(int i=triangle.size()-1;i>=0;i--){
        for(int j=0;j<triangle.get(i).size();j++){
            A[j] = Math.min(A[j],A[j+1])+triangle.get(i).get(j);
        }
    }
    return A[0];
}