0018. 4Sum (M)
4Sum (M)
题目
Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b+ c + 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]
]
题意
求四元组使其之和等于指定值。
思路
仿照 0015. 3Sum 的方法：先对数组排序，依次固定前两个元素i、j，对后两个元素m、n使用two pointers，注意去除重复值，时间复杂度为\(O(N^3)\)。
另一种方法：先预处理两个数的和，存入hash表中，再重新二重循环求两数的和sum，看能否在hash表中找到值为target-sum的key，理论上复杂度为\(O(N^2)\)，但考虑到去重步骤，实际时间还会增加。
代码实现
Java
Two Pointers
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> ans = new ArrayList<>();
Arrays.sort(nums);
for (int i = 0; i < nums.length - 3; i++) {
// 去除重复的i
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
for (int j = i + 1; j < nums.length - 2; j++) {
// 去除重复的j
if (j > i + 1 && nums[j] == nums[j - 1]) {
continue;
}
int m = j + 1, n = nums.length - 1;
while (m < n) {
int sum = nums[i] + nums[j] + nums[m] + nums[n];
if (sum < target) {
m++;
while (m < n && nums[m] == nums[m - 1]) {
m++;
}
} else if (sum > target) {
n--;
while (m < n && nums[n] == nums[n + 1]) {
n--;
}
} else {
List<Integer> quaternion = new ArrayList<>();
quaternion.add(nums[i]);
quaternion.add(nums[j]);
quaternion.add(nums[m]);
quaternion.add(nums[n]);
ans.add(quaternion);
m++;
while (m < n && nums[m] == nums[m - 1]) {
m++;
}
n--;
while (m < n && nums[n] == nums[n + 1]) {
n--;
}
}
}
}
}
return ans;
}
}
Hash
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> ans = new ArrayList<>();
Map<Integer, List<Pair>> hash = new HashMap<>();
Arrays.sort(nums);
// 生成hash表
for (int i = 0; i < nums.length - 1; i++) {
for (int j = i + 1; j < nums.length; j++) {
// 若hash中存在
if (hash.containsKey(nums[i] + nums[j])) {
boolean flag = true;
// 同元素组成的sum只保留x最大的下标对
for (Pair pair : hash.get(nums[i] + nums[j])) {
if (nums[pair.x] == nums[i]) {
pair.x = i;
pair.y = j;
flag = false;
break;
}
}
if (flag) {
hash.get(nums[i] + nums[j]).add(new Pair(i, j));
}
} else {
List<Pair> pair = new ArrayList<>();
pair.add(new Pair(i, j));
hash.put(nums[i] + nums[j], pair);
}
}
}
for (int i = 0; i < nums.length - 3; i++) {
// 去除重复的i
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
for (int j = i + 1; j < nums.length - 2; j++) {
// 去除重复的j
if (j > i + 1 && nums[j] == nums[j - 1]) {
continue;
}
int sum = nums[i] + nums[j];
if (hash.containsKey(target - sum)) {
for (Pair pair : hash.get(target - sum)) {
// 只记录下标排在j后面的元素
if (pair.x > j) {
List<Integer> quaternion = Arrays.asList(nums[i], nums[j], nums[pair.x], nums[pair.y]);
ans.add(quaternion);
}
}
}
}
}
return ans;
}
private class Pair {
int x, y;
public Pair(int x, int y) {
this.x = x;
this.y = y;
}
}
}
JavaScript
Two Pointers
/**
* @param {number[]} nums
* @param {number} target
* @return {number[][]}
*/
var fourSum = function (nums, target) {
let res = []
nums.sort((a, b) => a - b)
for (let i = 0; i < nums.length - 3; i++) {
if (i > 0 && nums[i] === nums[i - 1]) continue
for (let j = i + 1; j < nums.length - 2; j++) {
if (j > i + 1 && nums[j] === nums[j - 1]) continue
let k = j + 1, m = nums.length - 1
while (k < m) {
let sum = nums[i] + nums[j] + nums[k] + nums[m]
if ((k > j + 1 && nums[k] === nums[k - 1]) || sum < target) {
k++
} else if ((m < nums.length - 1 && nums[m] === nums[m + 1]) || sum > target) {
m--
} else {
res.push([nums[i], nums[j], nums[k++], nums[m--]])
}
}
}
}
return res
}
Hash
/**
* @param {number[]} nums
* @param {number} target
* @return {number[][]}
*/
var fourSum = function (nums, target) {
let res = []
let map = new Map()
nums.sort((a, b) => a - b)
for (let i = 0; i < nums.length - 1; i++) {
for (let j = i + 1; j < nums.length; j++) {
let sum = nums[i] + nums[j]
if (map.has(sum)) {
let pair = map.get(sum).find((pair) => nums[pair[0]] == nums[i])
if (pair) {
[pair[0], pair[1]] = [i, j]
} else {
map.get(sum).push([i, j])
}
} else {
map.set(sum, [[i, j]])
}
}
}
for (let i = 0; i < nums.length - 1; i++) {
if (i > 0 && nums[i] === nums[i - 1]) continue
for (let j = i + 1; j < nums.length; j++) {
if (j > i + 1 && nums[j] === nums[j - 1]) continue
let sum = nums[i] + nums[j]
if (map.has(target - sum)) {
for (let pair of map.get(target - sum)) {
if (pair[0] > j) {
res.push([nums[pair[0]], nums[pair[1]], nums[i], nums[j]])
}
}
}
}
}
return res
}
