Description

	#120	Triangle	Medium
		Dynamic Programming	Top 150
Given a triangle array, return the minimum path sum from top to bottom.

For each step, you may move to an adjacent number of the row below. More formally, if you are on index i on the current row, you may move to either index i or index i + 1 on the next row.

Example 1:

Input: triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]
Output: 11

Explanation: The triangle looks like:

The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above).

Example 2:

Input: triangle = [[-10]]
Output: -10

Solution

2D DP

It works, but not good, especially memory.

Runtime 3ms

Memory 44.4MB

class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int maxRowSize = triangle.get(triangle.size() -1).size();
        // create an array nxn to store the minimal distance that starts from triangle[i][j]
        int[][] dp = new int[maxRowSize][maxRowSize];

        for(int i = triangle.size() -1; i >=0; i--) {
            List<Integer> row = triangle.get(i);
            for(int j = 0; j < row.size(); j++) {
                if(i < dp.length -1) {
                    // calculate minimal distance and update
                    dp[i][j] = Math.min(dp[i+1][j], dp[i+1][j+1]) + row.get(j);
                } else {
                    // for the bottom line, there's no subpaths
                    dp[i][j] = row.get(j);
                }
            }
        }
        return dp[0][0];
    }
}

1D DP

Runtime 2ms

Memory 43.61MB

class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        // 可以直接用 triangle.size();
        int maxRowSize = triangle.get(triangle.size() -1).size();
        // 只需要记录上一次的最短路径
        int[] dp = new int[maxRowSize];

        for(int i = triangle.size() -1; i >=0; i--) {
            List<Integer> row = triangle.get(i);
            for(int j = 0; j < row.size(); j++) {
                if(i < dp.length -1) {
                    // 从左到右更新，正好计算右边的元素时，当前的值已经不需要用到了。所以可以直接更新
                    dp[j] = Math.min(dp[j], dp[j+1]) + row.get(j);
                } else {
                    dp[j] = row.get(j);
                }
            }
        }
        return dp[0];
    }
}

2D DP +DFS

Runtime 1ms

Memory 44.49MB

class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int[][]dp = new int[triangle.size()][triangle.size()];
        for(int[]temp: dp)
            Arrays.fill(temp, Integer.MAX_VALUE);
    
        return find(triangle, 0, 0, dp);    
    }
    private int find(List<List<Integer>>triangle, int i, int j, int[][]dp) {
        if(i == triangle.size())
            return 0;

        if(dp[i][j] != Integer.MAX_VALUE) 
            return dp[i][j];
        int current = triangle.get(i).get(j);
        int a = current + find(triangle, i+1, j, dp);
        int b = current + find(triangle, i+1, j+1, dp);
        return dp[i][j] = Math.min(a, b);
    }
}