Triangle 解答

Question

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).

Solution 1 -- DP

We define dp[i] to be the smallest sum that must include nums[i]. For easier understanding, we maintain two lists to record dp[i] information. Time complexity O(n^2) and space cost O(n).

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        if (triangle == null || triangle.size() < 1)
            return 0;
        int size = triangle.size(), result = Integer.MAX_VALUE;
        List<Integer> currentList, currentDP, prevDP = new ArrayList<Integer>();
        for (int i = 0; i < size; i++) {
            currentList = triangle.get(i);
            currentDP = new ArrayList<Integer>();
            if (i == 0) {
                currentDP.add(currentList.get(i));
            } else {
                for (int j = 0; j <= i; j++) {
                    int tmpMin;
                    // Three Cases
                    if (j == 0)
                        tmpMin = currentList.get(j) + prevDP.get(0);
                    else if (j == i)
                        tmpMin = currentList.get(j) + prevDP.get(j - 1);
                    else
                        tmpMin = currentList.get(j) + Math.min(prevDP.get(j), prevDP.get(j - 1));
                    currentDP.add(tmpMin);
                }
            }
            prevDP = currentDP;
        }
        // Select minimum number of dp[i]
        for (int tmp : prevDP)
            result = Math.min(tmp, result);
        return result;
    }
}

Solution 2 -- Bottom Up

In this way, we need not to consider the three cases discussed above.

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