216 Combination Sum III [Leetcode]

题目内容：

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Ensure that numbers within the set are sorted in ascending order.

Example 1:

Input: k = 3, n = 7

Output:

[[1,2,4]]

Example 2:

Input: k = 3, n = 9

Output:

[[1,2,6], [1,3,5], [2,3,4]]

解题思路：
保存一个目标对象数组，这里是1-9，然后使用回溯的方法求所有可能相加的结果。

代码实现如下：

class Solution {private:    vector<vector<int>> result;    vector<int> nums;public:    vector<vector<int>> combinationSum3(int k, int n) {        vector<int> mid_rlt;        nums = vector<int>(9);        for(int i = 1; i < 10; ++i)            nums[i-1] = i;        backtracking(mid_rlt, n, 0, k);        return result;    }    void backtracking(vector<int> mid_rlt, int remain, int index, int count) {        if(remain == 0 && count == 0) {            result.push_back(mid_rlt);            return;        }        for(int i = index; nums[i] <= remain && i < 9; ++i) {            mid_rlt.push_back(nums[i]);            backtracking(mid_rlt, remain - nums[i], i+1, count-1);            mid_rlt.pop_back();        }    }};

在上述基础上我们还可以进行剪枝。给定一个数字以及要求组成他的数字的个数，我们能够判断这些数字的个数能够组成的最大值和最小值。因此在回溯之前加上这样一个判断可以大大提高效率。

代码如下：

class Solution {private:    vector<vector<int>> result;    vector<int> nums;public:    vector<vector<int>> combinationSum3(int k, int n) {        vector<int> mid_rlt;        nums = vector<int>(9);        for(int i = 1; i < 10; ++i)            nums[i-1] = i;        backtracking(mid_rlt, n, 0, k);        return result;    }    void backtracking(vector<int> mid_rlt, int remain, int index, int count) {        if(getMin(count) > remain || getMax(count) < remain)            return;        if(count == 0) {            if(remain == 0)                result.push_back(mid_rlt);            return;        }        for(int i = index; nums[i] <= remain && i < 9; ++i) {            mid_rlt.push_back(nums[i]);            backtracking(mid_rlt, remain - nums[i], i+1, count-1);            mid_rlt.pop_back();        }    }    int getMin(int k) {        int minValue(0);        for(int i = 0; i < k; ++i)            minValue += nums[i];        return minValue;    }    int getMax(int k) {        int maxValue(0);        for(int i = 0; i < k; ++i)            maxValue += nums[8-i];        return maxValue;    }};