Triangle leetcode

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).

自底向上求解,

state:dp[i][j] 表示自底向上到第i行第j个数的最小路径

initial: dp[n][i] = triangle最后一个数组的值

function:dp[i][j] = Math.min(dp[i+1][j], dp[i+1][j+1]) + triangle.get(i).get(j) 最小值由当前点的值加上下一行相邻两个路径值最小的路径

return: dp[0][0]
time : O(n^2)

//O(n^2)space

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 = z; j >= 0; j--) {
                dp[i][j] = Math.min(dp[i+1][j], dp[i+1][j+1]) + triangle.get(i).get(j);
            }
        }
        return dp[0][0];
    }
}
//O(n) space

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];
        for (int i = 0; i < n; i++) {
           dp[i] = triangle.get(n - 1).get(i); 
        }
        for (int i = n-2; i>= 0; i--) {
            for (int j =0; j <= i; j++) {
                dp[j] = Math.min(dp[j], dp[j + 1]) + triangle.get(i).get(j);
            }
        }
        return dp[0];
    }
}