377. Combination Sum IV

标签：and   div   public   add   with   not   hang   等于   i++   

Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

Example:

nums = [1, 2, 3]
target = 4
The possible combination ways are:
(1, 1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(1, 3)
(2, 1, 1)
(2, 2)
(3, 1)
Note that different sequences are counted as different combinations.
Therefore the output is 7.

 Follow up:

What if negative numbers are allowed in the given array?
How does it change the problem?
What limitation we need to add to the question to allow negative numbers?

题目含义：给定多个正整数且没有重复，从中挑选任意个数字（可以重复）使其加起来等于给定目标值target

思路：

 dp[i]表示总和为i的情况数，
 假设当前整数值等于n(n小于i)的，如果我们知道了dp[i-n]的情况数，dp[i-n]的每一种情况加n都等于dp[i]，也就是说dp[i-n]和dp[i]的情况数相同。
 每获取一个整数n，都会给了dp[i]添加dp[i-n]种情况，所以公式为dp[i] = dp[i] + dp[i-n]

例如：
 当前元素n等于1时，dp[9] = dp[9] + dp[9-1]
 dp[8] = dp[8] + dp[8-1]
 ...
 dp[1] = dp[1] + dp[1-1]
 当前元素n等于2时，dp[9] = dp[9] + dp[9-2]
 dp[8] = dp[8] + dp[8-2]
 ...
 dp[2] = dp[2] + dp[2-2]
 当前元素n等于3时，dp[9] = dp[9] + dp[9-3]
 dp[8] = dp[8] + dp[8-3]
 ...
 dp[3] = dp[3] + dp[3-3]

 1     public int combinationSum4(int[] nums, int target) {                  
 2         int[] dp=new int[target+1];
 3         dp[0]=1;
 4         for(int i=1;i<=target;i++)
 5         {
 6             for (int num:nums)
 7             {
 8                 if(i>=num) dp[i]=dp[i]+dp[i-num];
 9             }
10         }
11         return dp[target];
12     }

377. Combination Sum IV

标签：and   div   public   add   with   not   hang   等于   i++   

原文地址：http://www.cnblogs.com/wzj4858/p/7694733.html