[LeetCode] 40. Combination Sum II ☆☆☆(数组相加等于指定的数)

https://leetcode.wang/leetCode-40-Combination-Sum-II.html

描述

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates 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.
Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
[1,2,2],
[5]
]

解析

回溯法

和上一道题非常像了，区别在于这里给的数组中有重复的数字，每个数字只能使用一次，然后同样是给出所有和等于 target 的情况。

代码

回溯法

public List<List<Integer>> combinationSum2(int[] candidates, int target) {
    List<List<Integer>> ans = new ArrayList<>();
    getAnswer(ans, new ArrayList<>(), candidates, target, 0);
    /*************修改的地方*******************/
    // 如果是 Input: candidates = [2,5,2,1,2], target = 5,
    // 输出会出现 [2 2 1] [2 1 2] 这样的情况，所以要去重
    return removeDuplicate(ans);
     /****************************************/
}

private void getAnswer(List<List<Integer>> ans, ArrayList<Integer> temp, int[] candidates, int target, int start) {
    if (target == 0) {
        ans.add(new ArrayList<Integer>(temp));
    } else if (target < 0) {
        return;
    } else {
        for (int i = start; i < candidates.length; i++) {
            temp.add(candidates[i]);
            /*************修改的地方*******************/
            //i -> i + 1 ，因为每个数字只能用一次，所以下次遍历的时候不从自己开始
            getAnswer(ans, temp, candidates, target - candidates[i], i + 1);
            /****************************************/
            temp.remove(temp.size() - 1);
        }
    }

}

private List<List<Integer>> removeDuplicate(List<List<Integer>> list) {
    Map<String, String> ans = new HashMap<String, String>();
    for (int i = 0; i < list.size(); i++) {
        List<Integer> l = list.get(i);
        Collections.sort(l);
        String key = "";
        for (int j = 0; j < l.size() - 1; j++) {
            key = key + l.get(j) + ",";
        }
        key = key + l.get(l.size() - 1);
        ans.put(key, "");
    }
    List<List<Integer>> ans_list = new ArrayList<List<Integer>>();
    for (String k : ans.keySet()) {
        String[] l = k.split(",");
        List<Integer> temp = new ArrayList<Integer>();
        for (int i = 0; i < l.length; i++) {
            int c = Integer.parseInt(l[i]);
            temp.add(c);
        }
        ans_list.add(temp);
    }
    return ans_list;
}

先排序回溯法

public List<List<Integer>> combinationSum2(int[] candidates, int target) {
    List<List<Integer>> ans = new ArrayList<>();
    Arrays.sort(candidates); //排序
    getAnswer(ans, new ArrayList<>(), candidates, target, 0);
    return ans;
}

private void getAnswer(List<List<Integer>> ans, ArrayList<Integer> temp, int[] candidates, int target, int start) {
    if (target == 0) {
        ans.add(new ArrayList<Integer>(temp));
    } else if (target < 0) {
        return;
    } else {
        for (int i = start; i < candidates.length; i++) {
            //跳过重复的数字
            if(i > start && candidates[i] == candidates[i-1]) continue;
            temp.add(candidates[i]);
            /*************修改的地方*******************/
            //i -> i + 1 ，因为每个数字只能用一次，所以下次遍历的时候不从自己开始
            getAnswer(ans, temp, candidates, target - candidates[i], i + 1);
            /****************************************/
            temp.remove(temp.size() - 1);
        }
    }
}