120. Triangle.md

120. Triangle

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).

Thinking:

• Method 1:dp:二维数组

class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        if(triangle == null || triangle.size() == 0) return 0;
        int len = triangle.size();
        int[][] dp = new int[len][len];
        dp[0][0] = triangle.get(0).get(0);
        for(int i = 1; i < len; 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 = 2; i < len; 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 result = dp[len - 1][0];
        for(int i = 1; i < len; i++){
            result = Math.min(result, dp[len - 1][i]);
        }
        return result;
    }
}

• Method 2: DP,一维，倒三角，从下往上，dp存储当前行的每一个位置的最短路径。

class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int len = triangle.size();
        int[] dp = new int[len + 1];
        for(int i = len - 1; i >= 0; i--){
            List<Integer> temp = triangle.get(i);
            for(int j = 0; j < temp.size(); j++){
                dp[j] = Math.min(dp[j], dp[j + 1]) + temp.get(j);
            }
        }
        return dp[0];
    }
}

二刷

和一刷的思路一样。使用一维数组dp.

class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int size = triangle.size();
        int[] dp = new int[size + 1];
        for(int i = size - 1; i >= 0; i--){
            List<Integer> temp = triangle.get(i);
            int sz = temp.size();
            for(int j = 0 ; j < sz; j++)
                dp[j] = Math.min(dp[j], dp[j + 1]) + temp.get(j);
        }
        return dp[0];
    }
}

Third time

• Method 1: dp + rotation array

   class Solution {
   	public int minimumTotal(List<List<Integer>> triangle) {
   		int height = triangle.size(), width = triangle.get(height - 1).size();
   		int[] dp = new int[width];
   		int temp = 0, pre = 0;
   		for(int i = 0; i < width; i++){
   			dp[i] = triangle.get(height - 1).get(i);
   		}
   		for(int i = height - 2; i >= 0; i--){
   			temp = dp[i + 1];
   			for(int j = i; j >= 0; j--){
   				pre = dp[j];
   				dp[j] = triangle.get(i).get(j) + Math.min(dp[j], temp);
   				temp = pre;
   			}
   		}
   		return dp[0];
   	}
   }