Given an array nums of n integers and an integer target, are there elements abc, and d in nums such that a + bc + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

The solution set must not contain duplicate quadruplets.

Example:

Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.
A solution set is:
[
[-1, 0, 0, 1],
[-2, -1, 1, 2],
[-2, 0, 0, 2]
]
 public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> res = new ArrayList<>();
Arrays.sort(nums);
for (int i = 0; i < nums.length - 3; i++) {
if (i != 0 && nums[i] == nums[i - 1])
continue;
for (int j = i + 1; j < nums.length - 2; j++) {
if (j > i + 1 && nums[j] == nums[j - 1])
continue;
int k = j + 1;
int l = nums.length - 1;
while (k < l) {
int sum = nums[i] + nums[j] + nums[k] + nums[l];
if (sum == target) {
List<Integer> list = new ArrayList<>();
list.add(nums[i]);
list.add(nums[j]);
list.add(nums[k]);
list.add(nums[l]);
res.add(list);
k++;
l--;
// 去重复
while (k < l && nums[k] == nums[k - 1]) {
k++;
}
while (k < l && nums[l] == nums[l + 1]) {
l--;
}
} else if (sum < target) {
k++;
} else {
l--;
}
}
}
}
return res;
}
 public List<List<Integer>> fourSum3(int[] num, int target) {
ArrayList<List<Integer>> ans = new ArrayList<>();
if (num.length < 4)
return ans;
Arrays.sort(num);
for (int i = 0; i < num.length - 3; i++) {
if (num[i] + num[i + 1] + num[i + 2] + num[i + 3] > target)
break; // first candidate too large, search finished
if (num[i] + num[num.length - 1] + num[num.length - 2] + num[num.length - 3] < target)
continue; // first candidate too small
if (i > 0 && num[i] == num[i - 1])
continue; // prevents duplicate result in ans list
for (int j = i + 1; j < num.length - 2; j++) {
if (num[i] + num[j] + num[j + 1] + num[j + 2] > target)
break; // second candidate too large
if (num[i] + num[j] + num[num.length - 1] + num[num.length - 2] < target)
continue; // second candidate too small
if (j > i + 1 && num[j] == num[j - 1])
continue; // prevents duplicate results in ans list
int low = j + 1, high = num.length - 1;
while (low < high) {
int sum = num[i] + num[j] + num[low] + num[high];
if (sum == target) {
ans.add(Arrays.asList(num[i], num[j], num[low], num[high]));
while (low < high && num[low] == num[low + 1])
low++; // skipping over duplicate on low
while (low < high && num[high] == num[high - 1])
high--; // skipping over duplicate on high
low++;
high--;
}
// move window
else if (sum < target)
low++;
else
high--;
}
}
}
return ans;
}

LeetCode_18 4Sum的更多相关文章

  1. [LeetCode] 4Sum II 四数之和之二

    Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such t ...

  2. [LeetCode] 4Sum 四数之和

    Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = tar ...

  3. LeetCode:3Sum, 3Sum Closest, 4Sum

    3Sum Closest Given an array S of n integers, find three integers in S such that the sum is closest t ...

  4. 2016/10/28 很久没更了 leetcode解题 3sum问题进阶版4sum

    18. 4Sum Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c  ...

  5. No.018:4Sum

    问题: Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = ...

  6. 6.3Sum && 4Sum [ && K sum ] && 3Sum Closest

    3Sum Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find a ...

  7. 3Sum & 4Sum

    3 Sum Given an array S of n integers, are there elements a, b, c in Ssuch that a + b + c = 0? Find a ...

  8. 【leetcode】4Sum

    4Sum Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d  ...

  9. 2sum、3sum、4sum以及任意连续的数的和为sum、任意连续或者不连续的数的和为sum

    2sum 如果数组是无序的,先排序(n*logn),然后用两个指针i,j,各自指向数组的首尾两端,令i=0,j=n-1,然后i++,j--,逐次判断a[i]+a[j]?=sum,如果某一刻a[i]+a ...

随机推荐

  1. HDU 3308 线段树单点更新+区间查找最长连续子序列

    LCIS                                                              Time Limit: 6000/2000 MS (Java/Oth ...

  2. mysql中decimal的使用

    float,double,decimal区别 创建表test_float_double_decimal CREATE TABLE `test_float_double_decimal` ( `id` ...

  3. spark groupByKey().mapValues

    >>> rdd = sc.parallelize([("bone", 231), ("bone", 21213), ("jack&q ...

  4. android 添加手机短信,获取手机短信,删除手机短信和修改手机短信

    注意添加权限: <uses-permission android:name="android.permission.READ_SMS"></uses-permis ...

  5. IDEA 单元测试

    下载所需的两个 jar 包,下载地址:Download and Install · junit-team/junit4 Wiki · GitHub junit-4.12.jar hamcrest-co ...

  6. 【158】◀▶ Linux-Bash学习

    鸟哥的 Linux 私房菜      Linux 的 26 个命令      Shell 脚本教程      Linux 命令大全 目录——按文件顺序: echo:显示变量内容 printf:格式化输 ...

  7. 虚拟机安装cenos7后ifcfg看网卡无inet地址掩码等信息

    在虚拟机安装centos7,进入系统使用ifconfig命令时,只有lo网卡( 127.0.0.1的ip地址)和eno16777736网卡,而且此网卡没有inet地址.掩码等信息. 这时候查看/etc ...

  8. git clone ssh permissions are too open 解决方案。

    错误如图所示 方案如下 https://stackoverflow.com/questions/9270734/ssh-permissions-are-too-open-error

  9. linux学习之路6 Vi文本编辑器

    vim是vi的增强版本 vim拥有三种模式: 命令模式(常规模式) vim启动后,默认进入命令模式.任何模式都可以通过按esc键回到命令模式(可以多按几次.命令模式可以通过键入不同的命令完成选择.复制 ...

  10. SpringMVC实现Action的两种方式以及与Struts2的区别

    4.程序员写的Action可采用哪两种方式? 第一.实现Controller接口第二.继承自AbstractCommandController接口 5.springmvc与struts2的区别? 第一 ...