[LeetCode] 377. Combination Sum IV 组合之和 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?

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

 

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posted @ 2018-03-30 05:16  轻风舞动  阅读(446)  评论(0编辑  收藏  举报