40. Combination Sum II(midum, backtrack, 重要)

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 the 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]
]

这题用到 backtrack 方法, 需去重.

e.g.

A = [1 1 2 5 6 7 10], target = 8

正确输出应该是：

[[1,1,6],[1,2,5],[1,7],[2,6]]

难点在去重条件：
* 我写的，错误：　`if(i >= 1 && A[i] == A[i - 1]) continue;`
- 错误输出: `[[1,2,5],[1,7],[2,6]]`
* 人家写的，正确: `if ((i >= 1) && (A[i] == A[i - 1]) && (i > start)) continue;`

差别就是 i > start 条件，挺不容易的想出来的．

自己想法，自个代码(被人家修正^^):

// backtrack
// A = [1 1 2 5 6 7 10]
vector<vector<int>> combinationSum2(vector<int>& A, int ta) {
	sort(A.begin(), A.end());
	vector < vector<int> > res;
	vector<int> temp;
	backtrack(A, res, temp, ta, 0);
	return res;
}

void backtrack(vector<int>& A, vector<vector<int> >& res, vector<int> temp,
		int remain, int start) {
	if (remain < 0)
		return;
	else if (remain == 0)
		res.push_back(temp);
	else {
		for (int i = start; i < A.size(); i++) {

			// not correct: if(A[i] == A[i - 1] && i >= 1) continue;
			// (i > start) is hard think, but important.

			if ((i >= 1) && (A[i] == A[i - 1]) && (i > start))
				continue; // check duplicate combination

			temp.push_back(A[i]);
			backtrack(A, res, temp, remain - A[i], i + 1); // i + 1, next element
			temp.pop_back();
		}
	}
}