题目：
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]
链接： http://leetcode.com/problems/combination-sum-ii/
题解：
依然是一道DFS + Backtracking题目。 与之前不同的是每个数字只允许使用一次。所以在回溯的循环里当 i > pos时，假如之后又重复的，continue。我们依然使用candidates[i]，而且candidates[i + 1]假如等于candidates[i]， 会在DFS的下一个阶段被使用到。当然假如不用这个巧妙的条件也可以， 可以在 target == 0的时候判断res中是否contains list，这样的话运行速度会慢不少，但也可以AC.
Time Complexity - O()， Space Complexity - O()。 如何计算复杂度，智商捉急啊...
public class Solution {
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
List<List<Integer>> res = new ArrayList<>();
if(candidates == null || candidates.length == 0)
return res;
Arrays.sort(candidates);
ArrayList<Integer> list = new ArrayList<>();
dfs(res, list, candidates, target, 0);
return res;
}
private void dfs(List<List<Integer>> res, ArrayList<Integer> list, int[] candidates, int target, int pos) {
if(target == 0) {
res.add(new ArrayList<Integer>(list));
return;
}
if(pos >= candidates.length || target < 0)
return;
for(int i = pos; i < candidates.length; i++) {
if(i > pos && candidates[i] == candidates[i - 1]) //for i > pos, if duplicate,continue. we still use candidates[i]
continue;
list.add(candidates[i]);
dfs(res, list, candidates, target - candidates[i], i + 1);
list.remove(list.size() - 1);
}
}
}
二刷：
这里依然是用了跟上一题目很接近的方法。不同的地方在于，每个数字不可以被无限次。所以一个数只能一次，而且遇到重复数字我们要跳过。这样我们在for循环里要加入一条 - if (i > pos && candidates[i] == candidates[i - 1]) continue; 并且在DFS的时候每次
每次新的position = i + 1, 并不是上一题的position = i。
看到有discuss里有方法用array来做backtracking，速度beat 99%，以后也可以把list改成array，试一试这种方法。
Java:
public class Solution {
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
List<List<Integer>> res = new ArrayList<>();
if (candidates == null || candidates.length == 0) {
return res;
}
Arrays.sort(candidates);
List<Integer> comb = new ArrayList<>();
combinationSum2(res, comb, candidates, target, 0);
return res;
}
private void combinationSum2(List<List<Integer>> res, List<Integer> comb, int[] candidates, int target, int pos) {
if (target < 0) {
return;
} else if (target == 0) {
res.add(new ArrayList<>(comb));
}
for (int i = pos; i < candidates.length; i++) {
if (i > pos && candidates[i] == candidates[i - 1]) {
continue;
}
int num = candidates[i];
if (num > target) {
return;
}
comb.add(num);
combinationSum2(res, comb, candidates, target - num, i + 1);
comb.remove(comb.size() - 1);
}
}
}
三刷:
跟上题唯一不同就是递归时把控制position的变量从 i 变成了 i + 1，这样我们就不会对一个元素进行多次计算。
Java:
public class Solution {
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
List<List<Integer>> res = new ArrayList<>();
if (candidates == null) return res;
Arrays.sort(candidates);
findCombinations(res, new ArrayList<>(), candidates, target, 0);
return res;
}
private void findCombinations(List<List<Integer>> res, List<Integer> list, int[] candidates, int target, int pos) {
if (target < 0) return;
if (target == 0) {
res.add(new ArrayList<>(list));
return;
}
for (int i = pos; i < candidates.length; i++) {
if (candidates[i] > target) break;
if (i > pos && candidates[i] == candidates[i - 1]) continue;
list.add(candidates[i]);
findCombinations(res, list, candidates, target - candidates[i], i + 1);
list.remove(list.size() - 1);
}
}
}
Reference:
https://leetcode.com/submissions/detail/51501884/