leetcode解题之 Combination Sum java 版（组合求和）

39. Combination Sum

Given a set of candidate numbers (C)(without duplicates) and a target number (T), find all unique combinations inC where the candidate numbers sums to T.

The same repeated number may be chosen fromC unlimited number of times.

Note:

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

For example, given candidate set [2, 3, 6, 7] and target7,
A solution set is:

[
  [7],
  [2, 2, 3]
]

求在给定的数组中查找为指定的和，每个数字可以重复使用多次，避免重复或者进行优化，数组要排序。

采用深度优先搜索的思想+回溯(可以不排序，并去掉剪枝)

	public List> combinationSum(int[] candidates, int target) {
		List> res = new ArrayList>();
		List tmp = new ArrayList<>();
		// 排序可以避免重复，结果可以按照顺序输出
		Arrays.sort(candidates);
		dfsCore(res, 0, 0, tmp, candidates, target);
		return res;
	}

	private void dfsCore(List> res, 
			int curIdx, int sum, List tmp, int[] candidates,
			int target) {
		if (sum > target)
			return;
		if (sum == target) {
			res.add(new ArrayList(tmp));
			return;
		}
		for (int i = curIdx; i < candidates.length; i++) {
			// 剪枝，可以没有，目的为了优化，必须先排序
			if (target < candidates[i])
				return;
			sum += candidates[i];
			// 剪枝，可以没有，目的为了优化，必须先排序
			if (target < sum)
				return;

			tmp.add(candidates[i]);
			// 之所以不传i+1的原因是：
			// The same repeated number may be
			// chosen from C unlimited number of time
			dfsCore(res, i, sum, tmp, candidates, target);
			tmp.remove(tmp.size() - 1);
			// 回溯
			sum -= candidates[i];
		}
	}

40. Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations inC where the candidate numbers sums toT.

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 target8,
A solution set is:

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

与上体题不同的是有重复数字，并且每个数字仅能使用一次。

必须先排序，避免重复

public List> combinationSum2(int[] candidates, int target) {
		List> res = new ArrayList>();
		List tmp = new ArrayList<>();
		// 此题必须先排序
		Arrays.sort(candidates);
		dfsCore(res, 0, 0, tmp, candidates, target);
		return res;
	}

	private void dfsCore(List> res, int curIdx, int sum, List tmp, int[] candidates,

			int target) {

		if (sum > target)
			return;
		if (sum == target) {
			res.add(new ArrayList(tmp));
			return;
		}
		//i = curIdx往后走，避免重复
		for (int i = curIdx; i < candidates.length; i++) {
			// 如果此层，下一个数跟当前数相等，则直接跳过，
			if (i > curIdx && candidates[i] == candidates[i - 1])
																	
				continue;
			// 剪枝，可以没有，目的为了优化，必须先排序
			if (target < candidates[i])
				return;
			sum += candidates[i];
			// 剪枝，可以没有，目的为了优化，必须先排序
			if (target < sum)
				return;

			tmp.add(candidates[i]);
			// 传入i+1
			dfsCore(res, i + 1, sum, tmp, candidates, target);
			tmp.remove(tmp.size() - 1);
			// 回溯
			sum -= candidates[i];
		}
	}

216. Combination Sum III

377. Combination Sum IV

78. Subsets

90. Subsets II