Combination Sum I & II & IIIleetcode

lpy1990

Combination Sum I

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.
Elements in a combination (a1, a2, … , ak) must be in non-descending
order. (ie, a1 ≤ a2 ≤ … ≤ ak). 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]

回溯法

说明

做法是我们遍历所有的结果, 然后找出符合的答案. 每次遍历我们只找当前的数和后面的数来选取, 所以要先给数组排序.这样就不会漏下答案

复杂度

时间O(n!) 空间栈O(n)

代码

public List<List<Integer>> combinationSum(int[] candidates, int target) {
    List<List<Integer>> res = new ArrayList<List<Integer>>();
    if (candidates == null || candidates.length == 0 ) {
        return res;
    }
    Arrays.sort(candidates);
    List<Integer> tem = new ArrayList<Integer>();
    helper(res, tem, target, candidates, 0);
    return res;
}
public void helper(List<List<Integer>> res, List<Integer> tem, int target, int[] candidates, int pos) {
    if (target == 0) {
        res.add(new ArrayList<Integer>(tem));
        return;
    }
    if (target < 0) {
        return;
    }
    for (int i = pos; i < candidates.length; i++) {
        tem.add(candidates[i]);
        helper(res, tem, target - candidates[i], candidates, i);
        tem.remove(tem.size() - 1);
    }
}

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.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
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] 

回溯法

说明

类似subsetII, 对于递归过程中每次新选取的可以是和递归数组有相同的,但是后面的元素就不能相同了

复杂度

时间O(n!) 空间栈O(n)

代码

public List<List<Integer>> combinationSum2(int[] candidates, int target) {
    List<List<Integer>> res = new ArrayList<List<Integer>>();
    if (candidates == null || candidates.length == 0 ) {
        return res;
    }
    Arrays.sort(candidates);
    List<Integer> tem = new ArrayList<Integer>();
    helper(res, tem, target, candidates, 0);
    return res;
}
public void helper(List<List<Integer>> res, List<Integer> tem, int target, int[] candidates, int pos) {
    if (target == 0) {
        res.add(new ArrayList<Integer>(tem));
        return;
    }
    if (target < 0) {
        return;
    }
    for (int i = pos; i < candidates.length; i++) {
        if (i != pos && candidates[i] == candidates[i - 1]) {
            continue;
        }
        tem.add(candidates[i]);
        helper(res, tem, target - candidates[i], candidates, i + 1);
        tem.remove(tem.size() - 1);
    }
}

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.

Ensure that numbers within the set are sorted in ascending order.

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]]

回溯法

说明

这道题更加类似combinations那道题, 只是返回条件不仅仅是k数目相等, 递归得到的n也要相等才能返回.

代码

public List<List<Integer>> combinationSum3(int k, int n) {
    List<List<Integer>> res = new ArrayList<List<Integer>>();
    if (n <= 0 || k <= 0) {
        return res;
    }
    List<Integer> tem = new ArrayList<Integer>();
    helper(tem, res, 1, n, k);
    return res;
}
public void helper(List<Integer> tem, List<List<Integer>> res, int index, int n, int k) {
    if (tem.size() == k && n == 0) {
        res.add(new ArrayList<Integer>(tem));
        return;
    }
    if (tem.size() > k || n < 0) {
        return;
    }
    for (int i = index; i <= 9; i++) {
        tem.add(i);
        helper(tem, res, i + 1, n - i, k);
        tem.remove(tem.size() - 1);
    }
}
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