题目

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).

思路

动态规划的题目。

1. 递归
2. 二维数组保存dp[i][j]：到(i,j)位置时的最小值
3. 自底向上一维数组 dp[i] = i索引处的最小值

代码

• 递归

 public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
             int sum;
             sum = getResult(triangle, 0, 0);
             return sum;
           }

       

    public int getResult(ArrayList<ArrayList<Integer>> triangle, int l, int k) {
             int sum = triangle.get(l).get(k);
             if (l < triangle.size() - 1)
                sum = sum + Math.min(getResult(triangle, l + 1, k), getResult(triangle, l + 1, k + 1));
             return sum;

• 从上至下

public int minimumTotal(List<List<Integer>> triangle) {
        if (triangle.size() == 0) {
            return 0;
        }
        int[][] dp = new int[triangle.size()][triangle.get(triangle.size() - 1).size()];
        //dp[i][j]：在（i,j）的最小路径
        dp[0][0] = triangle.get(0).get(0);
        for (int i = 1; i < triangle.size(); i++) {
            for (int j = 0; j < triangle.get(i).size(); j++) {
                if (j == 0) {
                    dp[i][j] = dp[i - 1][0] + triangle.get(i).get(j);
                } else if (j == triangle.get(i).size() - 1) {
                    dp[i][j] = dp[i - 1][j - 1] + triangle.get(i).get(j);
                } else {
                    dp[i][j] = triangle.get(i).get(j) + Math.min(dp[i - 1][j], dp[i - 1][j - 1]);
                }
            }
        }

        int res = Integer.MAX_VALUE;
        //在dp最后一行比较一个最小路径
        for (int i = 0; i < dp[dp.length - 1].length; i++) {
            res = Math.min(res, dp[dp.length - 1][i]);
        }
        return res;
    }

• 自底向上

    public int minimumTotal(List<List<Integer>> triangle) {
        if (triangle.size() == 0) {
            return 0;
        }
        int[] dp = new int[triangle.size()];
        for (int i = 0; i < dp.length; i++) {
            dp[i] = triangle.get(triangle.size() - 1).get(i);
        }
        //自底向上
        //dp[i]：到索引i的最小路径 dp[i] = arr[i] + min(dp[i+1] ,dp[i])
        //得到dp[0]
        for (int i = dp.length - 2; i >= 0; i--) {
            List<Integer> row = triangle.get(i);
            for (int j = 0; j < row.size(); j++) {
                dp[j] = row.get(j) + Math.min(dp[j], dp[j + 1]);
            }
        }
        return dp[0];
    }