120. Triangle java solutions

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).

Note:
Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

一开始是按照自顶向下的方式来寻找最小值，边界处理有点麻烦，调了几次。

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

 

 

解法二采用自底向上：

 

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

 如果题中没有负数的，其实可以用贪心做的。（注意：下面代码是无法AC的）

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int len = triangle.size();
        int ans = triangle.get(0).get(0);
        int index = 0;
        for(int i = 1; i <len; i++){
            ans += Math.min(triangle.get(i).get(index),triangle.get(i).get(index+1));
            index = triangle.get(i).get(index) < triangle.get(i).get(index+1) ? index : index + 1;
        }
        return ans;
    }
}