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  • 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];
    }
    

      

     

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  • 原文地址:https://www.cnblogs.com/superzhaochao/p/6474919.html
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