C++实现LeetCode(40.组合之和之二) C++实现LeetCode(40.组合之和之二

[LeetCode] 40. Combination Sum II 组合之和之二 Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
Each number in candidates may only be used once in the combination.
Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

Example 1:

【C++实现LeetCode(40.组合之和之二)】Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
[1,2,2],
[5]
]

这道题跟之前那道 Combination Sum 本质没有区别，只需要改动一点点即可，之前那道题给定数组中的数字可以重复使用，而这道题不能重复使用，只需要在之前的基础上修改两个地方即可，首先在递归的 for 循环里加上 if (i > start && num[i] == num[i - 1]) continue; 这样可以防止 res 中出现重复项，然后就在递归调用 helper 里面的参数换成 i+1，这样就不会重复使用数组中的数字了，代码如下：

class Solution {public:vector> combinationSum2(vector& num, int target) {vector> res; vector out; sort(num.begin(), num.end()); helper(num, target, 0, out, res); return res; }void helper(vector& num, int target, int start, vector& out, vector>& res) {if (target < 0) return; if (target == 0) { res.push_back(out); return; }for (int i = start; i < num.size(); ++i) {if (i > start && num[i] == num[i - 1]) continue; out.push_back(num[i]); helper(num, target - num[i], i + 1, out, res); out.pop_back(); }}};

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