40. Combination Sum II
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/
40. Combination Sum II的更多相关文章
- [Leetcode][Python]40: Combination Sum II
# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 40: Combination Sum IIhttps://oj.leetco ...
- [array] leetcode - 40. Combination Sum II - Medium
leetcode - 40. Combination Sum II - Medium descrition Given a collection of candidate numbers (C) an ...
- leetcode 39. Combination Sum 、40. Combination Sum II 、216. Combination Sum III
39. Combination Sum 依旧与subsets问题相似,每次选择这个数是否参加到求和中 因为是可以重复的,所以每次递归还是在i上,如果不能重复,就可以变成i+1 class Soluti ...
- 【LeetCode】40. Combination Sum II (2 solutions)
Combination Sum II Given a collection of candidate numbers (C) and a target number (T), find all uni ...
- [LeetCode] 40. Combination Sum II 组合之和之二
Given a collection of candidate numbers (candidates) and a target number (target), find all unique c ...
- 【LeetCode题意分析&解答】40. Combination Sum II
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in ...
- LeetCode OJ 40. Combination Sum II
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in ...
- 【一天一道LeetCode】#40. Combination Sum II
一天一道LeetCode系列 (一)题目 Given a collection of candidate numbers (C) and a target number (T), find all u ...
- [leetcode]40. Combination Sum II组合之和之二
Given a collection of candidate numbers (candidates) and a target number (target), find all unique c ...
- 39. Combination Sum + 40. Combination Sum II + 216. Combination Sum III + 377. Combination Sum IV
▶ 给定一个数组 和一个目标值.从该数组中选出若干项(项数不定),使他们的和等于目标值. ▶ 36. 数组元素无重复 ● 代码,初版,19 ms .从底向上的动态规划,但是转移方程比较智障(将待求数分 ...
随机推荐
- 解决sublime text 2总是在新窗口中打开文件
在mac下不是很喜欢sublime text 2 总是在新窗口中打开文件,很麻烦,文件打多了,就会出现N多窗口,虽然可以直接打开当前目录可以解决,但有时候查看其它项目中的单个文件,就比较麻烦.百度一直 ...
- 人工智能起步-反向回馈神经网路算法(BP算法)
人工智能分为强人工,弱人工. 弱人工智能就包括我们常用的语音识别,图像识别等,或者为了某一个固定目标实现的人工算法,如:下围棋,游戏的AI,聊天机器人,阿尔法狗等. 强人工智能目前只是一个幻想,就是自 ...
- D3js
http://d3js.org http://blog.csdn.net/lzhlzz/article/details/27497157
- windows下SSH客户端远程访问Linux出现错误
- Ubuntu12.04更新源地址列表
1. 修改更新源 sudo gedit /etc/apt/sources.list 2. 比较全的更新源列表中一般都包含deb和对应的deb-src. deb和对应的deb-src一般都包含preci ...
- .net 科学类型相关问题
Q:如果我要把使用科学记数法表示的string转换为int又该如何呢? A:你可以通过把NumberStyles.AllowDecimalPoint | NumberStyles.AllowExpon ...
- C# WinForm开发系列 - ZedGraph
ZedGraph是用于创建任意数据的二维线型.条型.饼型图表的一个类库,也可以作为Windows窗体用户控件和Asp.Net网页控件.这个类库具有高度的适应性,几乎所有式样的图表都能够被创建.这个类库 ...
- [转载]Windows 7 IIS (HTTP Error 500.21 - Internal Server Error)解决
今天在测试网站的时候,在浏览器中输入http://localhost/时,发生如下错误: HTTP Error 500.21 - Internal Server Error Handler " ...
- ssh-add 报错 Could not open a connection to your authentication agent
ERROR: [root@testcentos01 ~]# ssh-add Could not open a connection to your authentication agent 在shel ...
- 设置HTTP header方式
一, Server Code JSP----> <%@ page language="java" contentType="text/html; charse ...