code
一开始是这样写的，但是发现有重复的现象，于是把数组排序了一下，在回溯过程中添加判断条件去重。
class Solution {
List<List<Integer>> res = new ArrayList<>();
boolean vis[];
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
vis = new boolean[candidates.length];
List<Integer> path = new ArrayList<>();
helper(0, target, candidates, path);
return res;
}
private void helper(int id, int target, int[] nums, List<Integer> path){
if (id == nums.length) return;
if (target == 0){
res.add(new ArrayList<>(path));
return;
}
helper(id + 1, target, nums, path);
if (target >= nums[id] && !vis[id]){
vis[id] = true;
path.add(nums[id]);
helper(id, target - nums[id], nums, path);
vis[id] = false;
path.remove(path.size() - 1);
}
}
}
AC code
如果出现与前面一个值相等并且前面一个值没有用到，说明选当前id的情况已经考虑过了，所以可以直接跳过。
class Solution {
List<List<Integer>> res = new ArrayList<>();
boolean vis[];
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
vis = new boolean[candidates.length];
Arrays.sort(candidates);
List<Integer> path = new ArrayList<>();
helper(0, target, candidates, path);
return res;
}
private void helper(int id, int target, int[] nums, List<Integer> path){
if (id == nums.length) return;
if (target == 0){
res.add(new ArrayList<>(path));
return;
}
if (id > 0 && nums[id] == nums[id - 1] && !vis[id - 1]){
helper(id + 1, target, nums, path);
return;
}
if (target >= nums[id] && !vis[id]){
vis[id] = true;
path.add(nums[id]);
helper(id, target - nums[id], nums, path);
vis[id] = false;
path.remove(path.size() - 1);
}
helper(id + 1, target, nums, path);
}
}