Triangle leetcode

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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).

bottom up 自下往上的

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;    }}

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