120. Triangle

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).

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.

链接： http://leetcode.com/problems/triangle/

题解：

自底向上dp。虽然是个简单题， 但自己现在也能写出一些比较简练的代码了，是进步，要肯定。晚上东方串店撸串去！

Time Complexity - O(n)，Space Complexity - O(1)。

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

 

二刷:

一般来说对于题目给出的数据结构比如Tree或者List<>不要轻易修改。我们另外建立一个数组来dp就好了。这里主要使用了自底向上的思路，先初始化dp数组为triangle的最后一行，再用转移方程dp[i] = triangle.get(i).get(j) + Math.min(dp[i], dp[i + 1])就好了。要注意边界条件。

Java:

Time Complexity - O(n)，Space Complexity - O(n)。

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

 

三刷:

Java:

in-place:

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int min = Integer.MAX_VALUE;
        if (triangle == null || triangle.size() == 0) return Integer.MAX_VALUE;
        for (int i = triangle.size() - 2; i >= 0; i--) {
            for (int j = 0; j < triangle.get(i).size(); j++) {
                triangle.get(i).set(j, triangle.get(i).get(j) + Math.min(triangle.get(i + 1).get(j), triangle.get(i + 1).get(j + 1)));
            }
        }
        return triangle.get(0).get(0);
    }
}

 

Use auxiliary list

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int min = Integer.MAX_VALUE;
        if (triangle == null || triangle.size() == 0) return Integer.MAX_VALUE;
        
        List<Integer> res = new ArrayList<>(triangle.get(triangle.size() - 1));
                
        for (int i = triangle.size() - 2; i >= 0; i--) {
            for (int j = 0; j < triangle.get(i).size(); j++) {
                res.set(j, triangle.get(i).get(j) + Math.min(res.get(j), res.get(j + 1)));
            }
        }
        return res.get(0);
    }
}

 

 

Reference:

https://leetcode.com/discuss/5337/dp-solution-for-triangle

https://leetcode.com/discuss/10131/my-java-version-solution-with-o-n-space-accepted

https://leetcode.com/discuss/23544/my-8-line-dp-java-code-4-meaningful-lines-with-o-1-space

https://leetcode.com/discuss/20296/bottom-up-5-line-c-solution