Leet code -- Combination Sum系列整理

Combination Sum I

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.
例如: [2, 3, 6, 7] and target 7

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

给定一个数组（元素无重复），和一个目标值，找到所有组合，加起来等于目标值。数组中的元素可以重复使用.
解法：

public class CombinationSum {
    public static List<List<Integer>> combinationSum(int[] candidates, int target){
        Arrays.sort(candidates);
        List<List<Integer>> result = new ArrayList<>();
        getResult(result, new ArrayList<Integer>(), candidates, target, 0);
        return result;
    }

    private static void getResult(List<List<Integer>> result, ArrayList<Integer> current, int[] candidates, int target,
            int start) {
        if(target<0)    return;
        if(target==0){
            result.add(new ArrayList<>(current));
            return;
        }
        for(int i = start; i<candidates.length && target >= candidates[i]; i++){
            current.add(candidates[i]);
            getResult(result, current, candidates, target-candidates[i], i);
            current.remove(current.size() - 1);
        }
    }
    public static void main(String[] args) {
        int[] nums = {2,3,6,7};
        System.out.println(combinationSum(nums, 7));
    }
}

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LC40. Combination Sum II

给定一个数组（元素可以有重复），和一个目标值，找到所有组合，加起来等于目标值。数组中的元素不能重复使用。
例子: [10, 1, 2, 7, 6, 1, 5] and target 8

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

解法：

/**
 * 要去重，注意边界，递归时候要加一
 */
public class CombinationSum2 {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> result = new ArrayList<>();
        Arrays.sort(candidates);
        dfs(result, new ArrayList<Integer>(), candidates, target, 0);
        return result;
    }
    
    private void dfs(List<List<Integer>> result, ArrayList<Integer> current, int[] candidates, int target, int start) {
        if(target < 0)  return;
        if(target == 0){
            result.add(new ArrayList<Integer>(current));
            return;
        }
        for(int i = start; i<candidates.length && target >= candidates[i]; i++){
            current.add(candidates[i]);
            dfs(result, current, candidates, target - candidates[i], i+1);    // 此处注意i+1,每个元素只能用一次，加一后在向下递归
            current.remove(current.size()-1);
            while(i < candidates.length - 1 && candidates[i] == candidates[i+1]) i++;    // 去重复（注意上面有i+1,这里要length-1，边界问题）
        }
    }
    public static void main(String[] args) {
        int[] array = {10, 1, 2, 7, 6, 1, 5};
        int target = 8;
        System.out.println(new CombinationSum2().combinationSum2(array, target));
    }
}

LC216. Combination Sum III

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]]

给定K和N，从1--9中这几个9个数字组合出来K个数，其和为N。1-9不能重复使用.

/**
 * 注意结束条件：size达到k值 并且 剩余值为0
 */
public class CombinationSum3 {
    public List<List<Integer>> combinationSum3(int k, int n) {
        List<List<Integer>> result = new ArrayList<>();
        dfs(result, new ArrayList<Integer>(), k, n, 1);
        return result;
    }
    private void dfs(List<List<Integer>> result, ArrayList<Integer> current, int k, int remainder, int start){
        if(current.size() == k && remainder == 0){ //size达到k值 并且 剩余值为0
            result.add(new ArrayList<>(current));
            return ;
        }
        for(int i = start; i<=9 && remainder >= i; i++){
            current.add(i);
            dfs(result, current, k, remainder - i, i+1); // 不重复，i+1
            current.remove(current.size() - 1);
        }
    }
    public static void main(String[] args) {
        System.out.println(new CombinationSum3().combinationSum3(3, 15));
    }
}

LC 377. Combination Sum IV

Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

Example: nums = [1, 2, 3], target = 4

The possible combination ways are:

(1, 1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(1, 3)
(2, 1, 1)
(2, 2)
(3, 1)

Note that different sequences are counted as different combinations.
Therefore the output is 7.

Follow up: What if negative numbers are allowed in the given array?
How does it change the problem? What limitation we need to add to the
question to allow negative numbers?

如果有负数，就不能让数组中的元素重复使用。

给定一个正整数数组（元素无重复），给定目标target，找出组合的个数，使得组合中元素的和等于target。数组元素可以重复使用.

public class CombinationSum4 {
    public int combinationSum4(int[] candidates, int target){
        List<List<Integer>> result = new ArrayList<>();
        dfs(result, new ArrayList<Integer>(), candidates, target, 0);
        return result.size();
    }
    
    private void dfs(List<List<Integer>> result, ArrayList<Integer> current, int[] candidates, int target, int start) {
        if(target < 0)    return;
        if(target == 0){
            result.add(new ArrayList<>(current));
            return;
        }
        for(int i = 0; i<candidates.length && target >= candidates[i]; i++){
            current.add(candidates[i]);
            dfs(result, current, candidates, target-candidates[i], i);
            current.remove(current.size() - 1);
        }
    }
    public static void main(String[] args) {
        int[] arr = {1,2,3};
        System.out.println(new CombinationSum4().combinationSum4(arr, 4));
    }
}

递归调用循环中，对于第一题修改i的起始位置即可：i = 0
但是TLE。递归深度太深。
所以这个方法是不行的。
需要使用DP。

public int combinationSum4(int[] candidates, int target){
    Arrays.sort(candidates);
    int[] dp = new int[target + 1];
    dp[0] = 1;
    for(int i = 1; i<dp.length; i++){
        for(int curr: candidates){
            if(curr > i)    break;
            dp[i] += dp[i - curr];
        }
    }
    return dp[target];
}

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面试题：修改版

有道面经题目是一个修改版，也是返回组合个数即可，但是加了条件：去掉重复。
上面的例子：nums = [1, 2, 3] target = 4 ，返回 7.
The possible combination ways are: (1, 1, 1, 1) (1, 1, 2) (1, 2, 1) (1, 3) (2, 1, 1) (2, 2) (3, 1)
这个题目要返回的是4，所有的组合是：(1, 1, 1, 1) (1, 1, 2) (1, 3) (2, 2) (3, 1)
变成第一题了：需要改变返回值，返回大小即可。

看一下这几个的区别，轻微的改动，产生的不同结果：
以第一题Combination Sum I为基础：

public class CombinationSum {
    public static List<List<Integer>> combinationSum(int[] candidates, int target){
        Arrays.sort(candidates);
        List<List<Integer>> result = new ArrayList<>();
        getResult(result, new ArrayList<Integer>(), candidates, target, 0);
        return result;
    }
    
    private static void getResult(List<List<Integer>> result, ArrayList<Integer> current, int[] candidates, int target,
            int start) {
        if(target < 0)    return; // 是有可能小于0的
        if(target == 0){
            result.add(new ArrayList<>(current)); // 此处注意
            return;
        }
        // 注意点1
        for(int i = start; i<candidates.length && target >= candidates[i]; i++){
            current.add(candidates[i]);
            getResult(result, current, candidates, target-candidates[i], i); // 注意点2
            current.remove(current.size() - 1);
        }
    }
    public static void main(String[] args) {
        int[] nums = {1,2,3};
        System.out.println(combinationSum(nums, 4));
    }
}

在上面的两个注意点上：
第一题：数组（元素无重复），数组中的元素可以重复使用。结果

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

如果第一处修改成 i = 0 结果变为：

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

如果第一处修改为 i = start 以及 第二处修改为 i+1 结果变为：

[[1, 3]]

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