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.
- 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] ]
分析
这里和39类似，都是使用DFS。
唯一需要注意的是：[1,1,2,5,6,7,10] 中，找到8
如果使用39类似的方法，会出现两个[1,7]。 因为有两个重复的1。
解决方法：
在每一次调用helper函数的时候（代表同一层遍历）如果遇到c[i] == c[i - 1]，比如已经把 index = 0 的1所有的可能DFS过了，那么再遇到 index = 1 的 1时候，就跳过，不进行DFS，因为之前的结果一定包含这次的结果。
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class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> results;
sort(candidates.begin(), candidates.end());
helper(results, vector<int>{}, candidates, target, 0);
return results;
}
void helper(vector<vector<int>> &results, vector<int> result, vector<int>& c, int target, int index) {
for (int i = index; i < c.size(); ++i) {
int t = target - c[i];
if (t < 0) {
return;
}
else if(i==index || c[i] != c[i - 1]){ // jump over duplicate results
result.push_back(c[i]);
if (t == 0) {
results.push_back(result);
}
else {
helper(results, result, c, t, i + 1);// i + 1: insure every element is used once
}
result.pop_back();
}
}
}
}; |
精简版
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class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> results;
sort(candidates.begin(), candidates.end());
helper(results, vector<int>{}, candidates, target, 0);
return results;
}
void helper(vector<vector<int>> &results, vector<int> result, vector<int>& c, int target, int index) {
if( target == 0){
results.push_back(result);
return;
}
for (int i = index; i < c.size() && target >= c[i]; ++i) {
if(i==index || c[i] != c[i - 1]){ // jump over duplicate results
result.push_back(c[i]);
helper(results, result, c, target - c[i], i + 1);// i + 1: insure every element is used once
result.pop_back();
}
}
}
}; |