377. Combination Sum IV

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?

最先想到的是dfs搜索，每次遍历nums搜索下一个数字，当前sum>target的时候返回，超时。

由dfs搜索的过程可以想到dp，dp[i]=dp[i-j1]+dp[i-j2]+....+dp[i-jn]+ (i存在于nums ? 1:0 )，其中j1,j2...jn是nums数组中比target小的数，举个例子，nums=[1,2,3],target=6,dp[1]=1,dp[2]=dp[1]+1,dp[3]=dp[2]+dp[1]+1,dp[4]=dp[3]+dp[2]+dp[1]

,dp[5]=dp[4]+dp[3]+dp[2],dp[6]=dp[5]+dp[4]+dp[3]，

这样有了状态转移方程就可以写dp啦。

为了避免例如[100000000,200000000,300000000] ，target=1400000000，这样的数据导致dp数组开不下导致溢出 (测试数据中并没有这种数据，直接开数组dp=new int[target+1]也能过），使用记忆优化搜索

public class Solution {
HashMap<Integer, Integer> hashmap;

public int combinationSum4(int[] nums, int target)
{
int len=nums.length;
Arrays.sort(nums);
hashmap=new HashMap<>(len);
search(nums, target);
return hashmap.get(target);
}

public int search(int[] nums,int target)
{
if(hashmap.containsKey(target))
return hashmap.get(target);

int index=Arrays.binarySearch(nums, target);
boolean exist=index>=0;
if(index<0)
index=-(index+1);

int sum=(exist?1:0);

for(int i=index-1;i>=0;i--)
{
int dis=target-nums[i];
sum+=search(nums, dis);
}

hashmap.put(target, sum);
return sum;
}
}