LeetCode-120.Triangle

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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.

 

使用动态规划 递推,时间复杂度为O(n)

 1 public int minimumTotal(List<List<Integer>> triangle) {//dp my
 2         if(0==triangle.size()){
 3             return 0;
 4         }
 5         List<Integer> result = new ArrayList<>();
 6         result.addAll(triangle.get(triangle.size()-1));
 7         for(int i=triangle.size()-2;i>=0;i--){
 8             for(int j=0;j<=i;j++){
 9                 int min = result.get(j)<result.get(j+1)?(result.get(j)+triangle.get(i).get(j)):(result.get(j+1)+triangle.get(i).get(j));
10                 result.set(j,min);
11             }
12         }
13         return result.get(0);
14     }

 可在原始数据上修改,这样空间复杂度为O(1)

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