Leetcode Combinations问题总结

@author stormma
@date 2017/11/30

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生命不息，奋斗不止！

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题目1

Given two integers n and k, return all possible combinations of k numbers out of 1 … n

Example
n = 4, k = 2

            [2,4],            [3,4],            [2,3],            [1,2],            [1,3],            [1,4]

题目意思很简单，就是从n个数(1-n)中取k个数的组合。简单的dfs+回溯即可。

代码实现

/**     * <p>Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.</p>     *     * <em>Example 1</em>     *     * if n = 4 and k = 2     *     * <code>     *       [2,4],     *       [3,4],     *       [2,3],     *       [1,2],     *       [1,3],     *       [1,4]     * </code>     */    static class Question1 {        public List<List<Integer>> combine(int n, int k) {            List<List<Integer>> ans = new ArrayList<>();            dfs(k, n, 1, ans, new LinkedList<>());            return ans;        }        /**         * @param k 表示从start开始取k个数的组合         * @param n          * @param start         * @param ans         * @param current 临时的结果         */        private void dfs(int k, int n, int start, List<List<Integer>> ans, List<Integer> current) {            // 此次搜索完成            if (k == 0) {                ans.add(new ArrayList<>(current));                return;            }            for (int i = start; i <= n; i++) {                current.add(i);                dfs(k - 1, n, i + 1, ans, current);                ((LinkedList) current).pollLast();            }        }        public static void main(String[] args) {            System.out.println(new Question1().combine(4, 2));        }    }

题目2 Combination Sum

Given a set of candidate numbers (C) (without duplicates) 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]]

题目分析

与题目一类似，这次是在数组中找出和 = target的组合，我们仿照题目1的解法来dfs+回溯解决此问题。

代码实现

class Solution {    public List<List<Integer>> combinationSum(int[] candidates, int target) {        Arrays.sort(candidates);        List<List<Integer>> ans = new ArrayList<>();        dfs(candidates, target, ans, new ArrayList<>(), 0);        return ans;    }    private void dfs(int[] candidates, int target, List<List<Integer>> ans, List<Integer> current, int start) {        if (target == 0) {            ans.add(new ArrayList<>(current));            return;        }        for (int i = start; i < candidates.length; i++) {            int num = candidates[i];            if (num > target) return;            current.add(num);            dfs(candidates, target - num, ans, current, i);            current.remove(current.size() - 1);        }    }}

题目3 Combination Sum II

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:

题目分析

这道题目有个限制，一个元素只能被使用一次，而且不能有重复，比如[1, 1, 7]和[1, 7, 1]就是重复的。那么说明如果数组为[1, 1, 7, 1]， target = 9，那么我们的答案理应只有[1, 1, 7]不能包含其他重复的。此题对于Combination Sum那个题目来说，难点是去重。偷懒的做法是直接使用Set来进行去重。

方法二，我们先进行排序， 如:

[1, 1, 1, 7]

我们开始搜索，第一次搜索我们可以得到[1, 1, 7]这个答案，第二次搜索从index = 1，[index = 1] = 1 ==上次开始的[index = 0] = 1。所有，这种情况我们应该跳过。按照这个，我们很容易写出我们的代码

代码实现

class Solution {    public List<List<Integer>> combinationSum2(int[] candidates, int target) {        List<List<Integer>> ans = new ArrayList<>();        Arrays.sort(candidates);        dfs(candidates, target, 0, ans, new ArrayList<>());        return ans;    }    private void dfs(int[] candidates, int target, int start, List<List<Integer>> ans, List<Integer> current) {        if (target == 0) {            ans.add(new ArrayList<>(current));            return;        }        for (int i = start; i < candidates.length; i++) {            int num = candidates[i];            if (num > target) return;            if (i > start && candidates[i] == candidates[i - 1]) continue;            current.add(num);            dfs(candidates, target - num, i + 1, ans, current);            current.remove(current.size() - 1);        }    }}