377. Combination Sum IV
- Total Accepted: 2547
- Total Submissions: 6581
- Difficulty: Medium
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 = 4The 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?
思路:DP,可以自顶向下,也可以自底向上。
假设values[i]表示的是i的被组合情况,显然有values[i]=values[i-nums[0]]+...+values[i-nums[n-1]](n=nums.size())。
注意冗余遍历的存在,例如下面的代码就没有去除冗余的遍历:
1 class Solution { 2 public: 3 int combinationSum4(vector & nums, int target) { 4 if(target<=0){ 5 return !target; 6 } 7 int i,n=nums.size(),res=0; 8 for(i=0;i
所以可以先申请内存,记录已经计算过的数的情况。
代码:
自底向上:
1 class Solution { 2 public: 3 int combinationSum4(vector & nums, int target) { 4 //局部域参数申请,未初始化时,可能不为0 5 //例如 6 //int *values=new int[target]; 7 //此时values数组的每个数的取值为任意值! 8 vector values(target+1); 9 values[0]=1;10 for(int i=1;i<=target;i++){11 for(int j=0;j=0){14 values[i]=values[i]+values[temp];15 }16 }17 }18 return values[target];19 }20 };
自顶向下:
values初始化为0不合适,因为values[i]==0可以意味着i不能被所给数组成,此时i已经被遍历过;也可以意味着i没有被遍历。所以设置values的初始值为0,还是不能避免大量重复的遍历操作。
1 class Solution { 2 public: 3 int combinationSum4(vector nums,vector & values,int target) { 4 if(target<=0){ 5 return !target; 6 } 7 if(values[target]==-1){ 8 values[target]=0; 9 for(int i=0;i