[LeetCode] 377. Combination Sum IV 组合之和之四

时间:2024-01-04 21:36:08

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.

这道题是组合之和系列的第四道,博主开始想当然的以为还是用递归来解,结果写出来发现 TLE 了,的确 OJ 给了一个 test case 为 [4,1,2] 32,这个结果是 39882198,用递归需要好几秒的运算时间,实在是不高效,估计这也是为啥只让返回一个总和,而不是返回所有情况,不然机子就爆了。而这道题的真正解法应该是用 DP 来做,解题思想有点像之前爬*的那道题 Climbing Stairs,这里需要一个一维数组 dp,其中 dp[i] 表示目标数为i的解的个数,然后从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]。这个也很好理解,比如说对于 [1,2,3] 4,这个例子,当计算 dp[3] 的时候,3可以拆分为 1+x,而x即为 dp[2],3也可以拆分为 2+x,此时x为 dp[1],3同样可以拆为 3+x,此时x为 dp[0],把所有的情况加起来就是组成3的所有情况了,参见代码如下:

解法一:

class Solution {
public:
int combinationSum4(vector<int>& nums, int target) {
vector<int> dp(target + );
dp[] = ;
for (int i = ; i <= target; ++i) {
for (auto a : nums) {
if (i >= a) dp[i] += dp[i - a];
}
}
return dp.back();
}
};

如果 target 远大于 nums 数组的个数的话,上面的算法可以做适当的优化,先给 nums 数组排个序,然后从1遍历到 target,对于i小于数组中的数字x时,直接 break 掉,因为后面的数更大,其余地方不变,参见代码如下:

解法二:

class Solution {
public:
int combinationSum4(vector<int>& nums, int target) {
vector<int> dp(target + );
dp[] = ;
sort(nums.begin(), nums.end());
for (int i = ; i <= target; ++i) {
for (auto a : nums) {
if (i < a) break;
dp[i] += dp[i - a];
}
}
return dp.back();
}
};

我们也可以使用递归+记忆数组的形式,不过这里的记忆数组用的是一个 HashMap。在递归函数中,首先判断若 target 小于0,直接返回0,若 target 等于0,则返回1。若当前 target 已经在 memo 中存在了,直接返回 memo 中的值。然后遍历 nums 中的所有数字,对每个数字都调用递归,不过此时的 target 要换成 target-nums[i],然后将返回值累加到结果 res 中即可,参见代码如下:

解法三:

class Solution {
public:
int combinationSum4(vector<int>& nums, int target) {
unordered_map<int, int> memo;
return helper(nums, target, memo);
}
int helper(vector<int>& nums, int target, unordered_map<int, int>& memo) {
if (target < ) return ;
if (target == ) return ;
if (memo.count(target)) return memo[target];
int res = , n = nums.size();
for (int i = ; i < n; ++i) {
res += helper(nums, target - nums[i], memo);
}
return memo[target] = res;
}
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/377

类似题目:

Combination Sum

Combination Sum II

Combination Sum III

参考资料:

https://leetcode.com/problems/combination-sum-iv/

https://leetcode.com/problems/combination-sum-iv/discuss/85079/My-3ms-Java-DP-solution

https://leetcode.com/problems/combination-sum-iv/discuss/85036/1ms-Java-DP-Solution-with-Detailed-Explanation

https://leetcode.com/problems/combination-sum-iv/discuss/85120/C%2B%2B-template-for-ALL-Combination-Problem-Set

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