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<List<Integer>> combinationSum(int[] candidates, int target) {List<List<Integer>> res = new ArrayList<List<Integer>>();List<Integer> 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<List<Integer>> res, int curIdx, int sum, List<Integer> tmp, int[] candidates,int target) {if (sum > target)return;if (sum == target) {res.add(new ArrayList<Integer>(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 timedfsCore(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<List<Integer>> combinationSum2(int[] candidates, int target) {List<List<Integer>> res = new ArrayList<List<Integer>>();List<Integer> 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<List<Integer>> res, int curIdx, int sum, List<Integer> tmp, int[] candidates,int target) {if (sum > target)return;if (sum == target) {res.add(new ArrayList<Integer>(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+1dfsCore(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