[leetcode] 回溯法 Combination Sum 系列问题

leetcde中回溯法：具体的方法，我认为手把手教你 < leetcode > 中的回溯算法——多一点套路这篇博文讲的很清晰。

下面是leetcode中Combination Sum 系列问题使用回溯法的问题

39 . Combination Sum
题目：
Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C 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 target 7,
A solution set is:
[
[7],
[2, 2, 3]
]

题意：这个题是找出数组中相加=target的所有组合。

思路：求解该问题应该使用回溯法。

具体代码：

public class Solution {    List<List<Integer>> result = new ArrayList<>();//保存最后结果    public List<List<Integer>> combinationSum(int[] candidates, int target) {        Arrays.sort(candidates);        if(target < candidates[0] || candidates.length == 0 || candidates == null){            return result;        }        List<Integer> list = new ArrayList<>();        backTracking(candidates, target, 0, list);        return result;    }    public void backTracking(int[] candicates, int target, int start, List<Integer> list){        if(target < 0){            return;        }else if(target == 0){            result.add(new ArrayList<>(list));        }else{            for(int i = start; i < candicates.length; i++){                list.add(candicates[i]);                backTracking(candicates, target - candicates[i], i, list);                list.remove(list.size() - 1);            }        }    }}

40 . Combination Sum II
题目：
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]
]

题意：找出所有数相加等于n的所有组合，注意每个数字只能用一次。
思路：使用回溯法。

代码如下：

public class Solution {    List<List<Integer>> result = new ArrayList<>();    public List<List<Integer>> combinationSum2(int[] candidates, int target) {        Arrays.sort(candidates);        if(candidates[0] > target || candidates.length == 0 || candidates == null){            return result;        }        List<Integer> list = new ArrayList<>();        backTracking(candidates, target, 0, list);        Set<List<Integer>> set = new HashSet<>(result);        return new ArrayList<>(set);    }    public void backTracking(int[] candidates, int target, int start, List<Integer> list){        if(target < 0){            return;        }else if(target == 0){            result.add(new ArrayList(list));        }else{            for(int i = start; i < candidates.length; i++){                list.add(candidates[i]);                backTracking(candidates, target - candidates[i], i + 1, list);                list.remove(list.size() - 1);            }        }    }}

216 . Combination Sum III

问题：
Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Example 1:

Input: k = 3, n = 7

Output:

[[1,2,4]]

Example 2:

Input: k = 3, n = 9

Output:

[[1,2,6], [1,3,5], [2,3,4]]

题意：找出k个数相加等于n的所有组合。
思路：使用回溯法。

具体代码：

public class Solution {    List<List<Integer>> result = new ArrayList<>();    public List<List<Integer>> combinationSum3(int k, int n) {        if(n < k ||n >= k * 9){            return result;        }        int[] candidates = new int[9];        for(int i = 0; i < 9; i++){            candidates[i] = i + 1;        }        List<Integer> list = new ArrayList<>();        backTracking(candidates,k, n, 0, list);        return result;    }    public void backTracking(int[] candidates, int k,int n, int start, List<Integer> list){        if(n < 0){            return;        }else if( n == 0){            if(list.size() == k)                result.add(new ArrayList<>(list));        }else{            int len = candidates.length;            for(int i = start; i < len; i++){                list.add(candidates[i]);                backTracking(candidates, k, n - candidates[i], i + 1, list);                list.remove(list.size() - 1);            }        }    }}