【LeetCode】Combination Sum I & II 解题报告

【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]

【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]

【解析】
给定一个数组，从中找出一组数来，使其和等于target。数组无序，但都是正整数。

I和II不同的是，I数组里没有重复的数，但一个数可以用多次；II数组里有重复，一个数只能用一次。

I和II都要求返回结果中没有重复的解，且每个解中的数都按非递减排好序。

方法：回溯。先对数组进行排序，然后从小到大累加，等于或超过target时回溯。

【Combination Sum I】

public class Solution {
    List<List<Integer>> ans = new ArrayList<List<Integer>>();
    int[] cans = {};
    
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        this.cans = candidates;
        Arrays.sort(cans);
        backTracking(new ArrayList(), 0, target);
        return ans;
    }
    
    public void backTracking(List<Integer> cur, int from, int target) {
        if (target == 0) {
            List<Integer> list = new ArrayList<Integer>(cur);
            ans.add(list);
        } else {
            for (int i = from; i < cans.length && cans[i] <= target; i++) {
                cur.add(cans[i]);
                backTracking(cur, i, target - cans[i]);
                cur.remove(new Integer(cans[i]));
            }
        }
    }
}

注意第19行代码，当加入cans[i]后，下一次还是从i开始，因为一个数可以用多次。（与下面代码40行区别）

【Combination Sum II 】

public class Solution {
    List<List<Integer>> ans = new ArrayList<List<Integer>>();
    int[] num = {};
    
    public List<List<Integer>> combinationSum2(int[] num, int target) {
        this.num = num;
        Arrays.sort(num);
        backTracking(new ArrayList<Integer>(), 0, target);
        return ans;
    }
    
    public void backTracking(List<Integer> cur, int from, int tar) {
        if (tar == 0) {
            //查看该解是否已经在结果集中，如对于输入[1,1]和1，只需放一个[1]到结果集中
            boolean exist = false;
            for (int i = ans.size() - 1; i >= 0 ; i--) {
                List<Integer> tmp = ans.get(i);
                if (tmp.size() != cur.size()) {
                    continue;
                }
                int j = 0;
                while (j < cur.size() && tmp.get(j) == cur.get(j)) {
                    j++;
                }
                if (j == cur.size()) {
                    exist = true;
                    break;
                }
            }
            
            //如果当前解不在结果集中，把其加入到结果集中
            if (!exist) {
                List<Integer> list = new ArrayList<Integer>(cur);
                ans.add(list);
            }
            return;
        }
        for (int i = from; i < num.length && num[i] <= tar; i++) {
            cur.add(num[i]);
            backTracking(cur, i + 1, tar - num[i]);
            cur.remove(new Integer(num[i]));
        }
    }
}

需要注意的是II的解法中，会出现相同的解，所以需要检查重复。（思考：I中为何不会出现重复解呢？）