题目描述
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.
解题思路
类似完全背包问题(可重复背包)
f[0]=1
f[i] 组成和为i的方案数
f[i]=求和(f[i-nums[j])
(对于示例中的数值:
最后一个为1时,求有多少个组成和为4-1=3的方案数;
最后一个为2时,求有多少个组成和为4-2=2的方案数;
最后一个为3时,求有多少个组成和为4-3=1的方案数;
即f[4]=f[4-1]+f[4-2]+f[4-3]=f[3]+f[2]+f[1]
)
return f[target]
target依次从1,2,3,4变化。
时间复杂度
O(target)*O(nums.size() ),即O(n*n)
代码如下:
class Solution {
public:
int combinationSum4(vector<int>& nums, int target) {
if(nums.empty()) return 0;
sort(nums.begin(),nums.end());
vector<int> dp(target+1,0);
for(int i = 1; i <= target; i++){
int count = 0;
for(int j = 0; j < nums.size(); j++){
if(i < nums[j]) break;
if(i == nums[j]) count++;
else count += dp[i-nums[j]];
}
dp[i] = count;
}
/*for (int i=0; i<dp.size(); i++) {
cout << dp[i] << endl;
}*/
return dp[target];
}
};