LeetCode-Triangle[dp]

标签：minimum   nbsp   i+1   bottom   poi   sum   logs   ges   get   

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.

思路：动态规划，设发f[i,j]表示从位置(i,j)出发到达三角形底部的最小路径和，所以状态转移方程有：

f[i,j]=min(f[i+1,j],f[i+1,j+1])    (i<size-1&&i>=0,j<i+1);

参考代码：

 1 public class Solution {
 2     public static int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
 3         int trianlgesize=triangle.size();
 4         int dp[][]=new int[trianlgesize][trianlgesize];
 5         for(int i=0;i<triangle.get(trianlgesize-1).size();i++){
 6             dp[trianlgesize-1][i]=triangle.get(trianlgesize).get(i);
 7         }
 8         for(int i=trianlgesize-2;i>=0;i--){
 9             for(int j=0;j<i+1;j++){
10                 dp[i][j]=Math.min(dp[i+1][j], dp[i+1][j+1])+triangle.get(i).get(j);
11             }
12         }
13         return dp[0][0];
14     }
15 }

LeetCode-Triangle[dp]

标签：minimum   nbsp   i+1   bottom   poi   sum   logs   ges   get   

原文地址：http://www.cnblogs.com/xiaolu266/p/7140759.html