leetcode 120. 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).

方法一

 bottom up 从下往上

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        if (triangle == null || triangle.size() == 0) {
            return 0;
        }
        int n = triangle.size();
        int[][]  dp = new int[n][n];
        
        for (int i = 0; i < n; i++) {
            dp[n - 1][i] = triangle.get(n-1).get(i);
        }
        for (int i = n - 2; i >= 0; i--) {
            for (int j = 0; j < triangle.get(i).size(); j++) {
                dp[i][j] = Math.min(dp[i + 1][j], dp[i + 1][j + 1]) + triangle.get(i).get(j);
            }
        }
        return dp[0][0];
        
    }
}

第二种方法，从上往下

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