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?
Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.
解法1：递归。按照前面I, II的思路用递归来解，会TLE，比如：OJ一个test case为[4,1,2] 32，结果是39882198，用递归需要好几秒时间。
解法2：动态规划DP来解。这道题类似于322. Coin Change ，建一个一维数组dp，dp[i]表示目标数target为i时解的个数，从1遍历到target，对于每一个数i，遍历nums数组，如果i>=x, dp[i] += dp[i - x]。比如[1,2,3] 4，当计算dp[3]时，3可以拆分为1+x，而x即为dp[2]，3也可以拆分为2+x，x为dp[1]，3同样可以拆为3+x，x为dp[0]，把所有情况加起来就是组成3的所有解。
Java: Recursive
public int combinationSum4(int[] nums, int target) {
if (target == 0) {
return 1;
}
int res = 0;
for (int i = 0; i < nums.length; i++) {
if (target >= nums[i]) {
res += combinationSum4(nums, target - nums[i]);
}
}
return res;
}
Java:
private int[] dp;
public int combinationSum4(int[] nums, int target) {
dp = new int[target + 1];
Arrays.fill(dp, -1);
dp[0] = 1;
return helper(nums, target);
}
private int helper(int[] nums, int target) {
if (dp[target] != -1) {
return dp[target];
}
int res = 0;
for (int i = 0; i < nums.length; i++) {
if (target >= nums[i]) {
res += helper(nums, target - nums[i]);
}
}
dp[target] = res;
return res;
}
Java:
public int combinationSum4(int[] nums, int target) {
int[] comb = new int[target + 1];
comb[0] = 1;
for (int i = 1; i < comb.length; i++) {
for (int j = 0; j < nums.length; j++) {
if (i - nums[j] >= 0) {
comb[i] += comb[i - nums[j]];
}
}
}
return comb[target];
}
Java：
class Solution {
public int combinationSum4(int[] nums, int target) {
if(nums==null || nums.length==0)
return 0;
int[] dp = new int[target+1];
dp[0]=1;
for(int i=0; i<=target; i++){
for(int num: nums){
if(i+num<=target){
dp[i+num]+=dp[i];
}
}
}
return dp[target];
}
}
Python:
class Solution(object):
def combinationSum4(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
dp = [0] * (target+1)
dp[0] = 1
nums.sort()
for i in xrange(1, target+1):
for j in xrange(len(nums)):
if nums[j] <= i:
dp[i] += dp[i - nums[j]]
else:
break
return dp[target]
Python:
class Solution(object):
def combinationSum4(self, nums, target):
nums, combs = sorted(nums), [1] + [0] * (target)
for i in range(target + 1):
for num in nums:
if num > i: break
if num == i: combs[i] += 1
if num < i: combs[i] += combs[i - num]
return combs[target]
C++:
class Solution {
public:
int combinationSum4(vector<int>& nums, int target) {
vector<int> dp(target + 1, 0);
dp[0] = 1;
sort(nums.begin(), nums.end());
for (int i = 1; i <= target; ++i) {
for (int j = 0; j < nums.size() && nums[j] <= i; ++j) {
dp[i] += dp[i - nums[j]];
}
}
return dp[target];
}
};
类似题目：
[LeetCode] 322. Coin Change 硬币找零
[LeetCode] 39. Combination Sum 组合之和
[LeetCode] 40. Combination Sum II 组合之和 II
[LeetCode] 216. Combination Sum III 组合之和 III