Triangle leetcode java

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.

 

题解：

一道动态规划的经典题目。需要自底向上求解。

递推公式是： dp[i][j] = dp[i+1][j] + dp[i+1][j+1] ，当前这个点的最小值，由他下面那一行临近的2个点的最小值与当前点的值相加得到。

由于是三角形，且历史数据只在计算最小值时应用一次，所以无需建立二维数组，每次更新1维数组值，最后那个值里存的就是最终结果。

代码如下：

 

 
   1  
  public  
  int minimumTotal(List<List<Integer>> triangle) { 
  
 
   2      
  if(triangle.size()==1) 
  
 
   3          
  return triangle.get(0).get(0); 
  
 
   4          
  
 
   5      
  int[] dp =  
  new  
  int[triangle.size()]; 
  
 
   6  
  
 
   7      
  // 
  initial by last row  
  
 
   8  
       
  for ( 
  int i = 0; i < triangle.get(triangle.size() - 1).size(); i++) { 
  
 
   9         dp[i] = triangle.get(triangle.size() - 1).get(i); 
  
 
  10     } 
  
 
  11   
  
 
  12      
  // 
   iterate from last second row 
  
 
  13  
       
  for ( 
  int i = triangle.size() - 2; i >= 0; i--) { 
  
 
  14          
  for ( 
  int j = 0; j < triangle.get(i).size(); j++) { 
  
 
  15             dp[j] = Math.min(dp[j], dp[j + 1]) + triangle.get(i).get(j); 
  
 
  16         } 
  
 
  17     } 
  
 
  18   
  
 
  19      
  return dp[0]; 
  
 
  20 } 
 

Reference:  http://www.programcreek.com/2013/01/leetcode-triangle-java/