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【LeetCode】Combination Sum II (2 solutions)
Combination Sum II
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
Each number in C may only be used once in the combination.
Note:
- All numbers (including target) will be positive integers.
- Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
- The solution set must not contain duplicate combinations.
For example, given candidate set 10,1,2,7,6,1,5 and target 8,
A solution set is: [1, 7] [1, 2, 5] [2, 6] [1, 1, 6]
解法一：
最直接的想法就是把所有子集都列出来，然后逐个计算和是否为target
但是考虑到空间复杂度，10个数的num数组就有210个子集，因此必须进行“剪枝”，去掉不可能的子集。
先对num进行排序。
在遍历子集的过程中：
（1）单个元素大于target，则后续元素无需扫描了，直接返回结果。
（2）单个子集元素和大于target，则不用加入当前的子集容器了。
（3）单个子集元素和等于target，加入结果数组。
class Solution {public: vector<vector<int> > combinationSum2(vector<int> &num, int target) { vector<vector<int> > result; vector<vector<int> > subsets; vector<int> empty; subsets.push_back(empty); sort(num.begin(), num.end()); for(int i = 0; i < num.size();) {//for each number int count = 0; int cur = num[i]; if(cur > target) //end return result; while(i < num.size() && num[i] == cur) {//repeat count i ++; count ++; } int size = subsets.size(); //orinigal size instead of calling size() function for(int j = 0; j < size; j ++) { vector<int> sub = subsets[j]; int tempCount = count; while(tempCount --) { sub.push_back(cur); int sum = accumulate(sub.begin(), sub.end(), 0); if(sum == target) { result.push_back(sub); subsets.push_back(sub); } else if(sum < target) subsets.push_back(sub); } } } return result; }};
解法二：递归回溯
需要注意的是：
1、在同一层递归树中，如果某元素已经处理并进入下一层递归，那么与该元素相同的值就应该跳过。否则将出现重复。
例如：1,1,2,3
如果第一个1已经处理并进入下一层递归1,2,3
那么第二个1就应该跳过，因为后续所有情况都已经被覆盖掉。
2、相同元素第一个进入下一层递归，而不是任意一个
例如：1,1,2,3
如果第一个1已经处理并进入下一层递归1,2,3，那么两个1是可以同时成为可行解的
而如果选择的是第二个1并进入下一层递归2,3，那么不会出现两个1的解了。
class Solution {public: vector<vector<int> > combinationSum2(vector<int> &num, int target) { vector<vector<int> > result; vector<int> cur; sort(num.begin(), num.end()); Helper(num, result, cur, 0, target); return result; } void Helper(vector<int> &num, vector<vector<int> > &result, vector<int> cur, int ind, int target) { int last; //first element for(int i = ind; i < num.size(); i ++) { if(i == ind) { cur.push_back(num[i]); int sum = accumulate(cur.begin(), cur.end(), 0); if(sum == target) { result.push_back(cur); return; } else if(sum < target) Helper(num, result, cur, i+1, target); else return; cur.pop_back(); last = num[i]; //start from last has been tried } else { //start from different element while(i < num.size() && num[i] == last) i ++; if(i == num.size()) return; if(num[i] > target) return; cur.push_back(num[i]); int sum = accumulate(cur.begin(), cur.end(), 0); if(sum == target) { result.push_back(cur); return; } else if(sum < target) Helper(num, result, cur, i+1, target); else return; cur.pop_back(); last = num[i]; //start from last has been tried } } }};
【LeetCode】Combination Sum II (2 solutions)