关于我的 Leetcode 题目解答，代码前往 Github：https://github.com/chenxiangcyr/leetcode-answers
LeetCode题目：1. Two Sum
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
以下代码的时间复杂度为 O(n)：
class Solution {
public int[] twoSum(int[] nums, int target) {
// key: integer, value: the idx for the integer
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int i = 0; i < nums.length; i++) {
int rest = target - nums[i];
if(map.containsKey(rest)) {
return new int[] {i, map.get(rest)};
}
map.put(nums[i], i);
}
throw new IllegalArgumentException("no result for the input");
}
}
LeetCode题目：167. Two Sum II - Input array is sorted
Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution and you may not use the same element twice.
Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2
以下代码的时间复杂度为 O(n)：
class Solution {
public int[] twoSum(int[] numbers, int target) {
int low = 0;
int high = numbers.length - 1;
while (low < high) {
int sum = numbers[low] + numbers[high];
if (sum == target) {
return new int[]{low + 1, high + 1};
}
else if (sum < target) {
low++;
}
else {
high--;
}
}
throw new IllegalArgumentException("no result for the input");
}
}
LeetCode题目：170. Two Sum III - Data structure design
Design and implement a TwoSum class. It should support the following operations: add and find.
-
add- Add the number to an internal data structure. -
find- Find if there exists any pair of numbers which sum is equal to the value.
For example,
add(1); add(3); add(5);
find(4) -> true
find(7) -> false
class TwoSum {
// 记录所有的数字
private List<Integer> list;
// 记录每个数字出现的次数
private Map<Integer, Integer> map;
/** Initialize your data structure here. */
public TwoSum() {
list = new ArrayList<Integer>();
map = new HashMap<Integer, Integer>();
}
/** Add the number to an internal data structure.. */
public void add(int number) {
if (map.containsKey(number)) {
map.put(number, map.get(number) + 1);
}
else {
map.put(number, 1);
list.add(number);
}
}
/** Find if there exists any pair of numbers which sum is equal to the value. */
public boolean find(int value) {
for (int i = 0; i < list.size(); i++){
int num1 = list.get(i);
int num2 = value - num1;
if ((num1 == num2 && map.get(num1) > 1) || (num1 != num2 && map.containsKey(num2))) {
return true;
}
}
return false;
}
}
/**
* Your TwoSum object will be instantiated and called as such:
* TwoSum obj = new TwoSum();
* obj.add(number);
* boolean param_2 = obj.find(value);
*/
LeetCode题目：15. 3Sum
Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note: The solution set must not contain duplicate triplets.
For example, given array S = [-1, 0, 1, 2, -1, -4],
A solution set is:
[-1, 0, 1],
[-1, -1, 2]
以下代码的时间复杂度为 O(n^2)：
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
return threeSum(nums, 0);
}
public List<List<Integer>> threeSum(int[] nums, int target) {
List<List<Integer>> result = new ArrayList<>();
// 排序
Arrays.sort(nums);
for(int i = 0; i < nums.length - 2; i++) {
// 避免重复数字
while(i > 0 && i < nums.length - 2 && nums[i] == nums[i - 1]) {
i++;
}
int low_idx = i + 1;
int high_idx = nums.length - 1;
int remaining = target - nums[i];
while(low_idx < high_idx) {
if(nums[low_idx] + nums[high_idx] < remaining) {
low_idx++;
}
else if (nums[low_idx] + nums[high_idx] > remaining) {
high_idx--;
}
else {
result.add(Arrays.asList(nums[i], nums[low_idx], nums[high_idx]));
// 避免重复数字
low_idx++;
while(low_idx < high_idx && nums[low_idx] == nums[low_idx - 1]) {
low_idx++;
}
// 避免重复数字
high_idx--;
while(low_idx < high_idx && nums[high_idx] == nums[high_idx + 1]) {
high_idx--;
}
}
}
}
return result;
}
}
LeetCode题目：16. 3Sum Closest
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
以下代码的时间复杂度为 O(n^2)：
class Solution {
public int threeSumClosest(int[] nums, int target) {
int result = nums[0] + nums[1] + nums[2];
// 排序
Arrays.sort(nums);
for(int i = 0; i < nums.length - 2; i++) {
int low_idx = i + 1;
int high_idx = nums.length - 1;
while(low_idx < high_idx) {
int sum = nums[i] + nums[low_idx] + nums[high_idx];
if (Math.abs(sum - target) < Math.abs(result - target)) {
result = sum;
}
if(sum < target) {
low_idx++;
}
else {
high_idx--;
}
}
}
return result;
}
}
LeetCode题目：18. 4Sum
Given an array S of n integers, are there elements a, b, c, and d in S 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.
For example, given array S = [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]]
以下代码的时间复杂度为 O(n^3)：
public class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
// 排序
Arrays.sort(nums);
// 转换为 k Sum 的问题
return kSum(nums, target, 4, 0, nums.length);
}
// 转换为 k Sum 的问题
private ArrayList<List<Integer>> kSum(int[] nums, int target, int k, int startIndex, int len) {
ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();
if(startIndex >= len) {
return res;
}
// 2 Sum 问题
if(k == 2) {
int low_idx = startIndex;
int high_idx = len - 1;
while(low_idx < high_idx) {
// 找到
if(nums[low_idx] + nums[high_idx] == target) {
List<Integer> temp = new ArrayList<>();
temp.add(nums[low_idx]);
temp.add(nums[high_idx]);
res.add(temp);
// 避免重复数字
low_idx++;
while(low_idx < high_idx && nums[low_idx] == nums[low_idx - 1]) {
low_idx++;
}
// 避免重复数字
high_idx--;
while(low_idx < high_idx && nums[high_idx] == nums[high_idx + 1]) {
high_idx--;
}
}
else if (nums[low_idx] + nums[high_idx] < target) {
low_idx++;
}
else {
high_idx--;
}
}
}
else {
for (int i = startIndex; i < len - k + 1; i++) {
// 递归为 k - 1 Sum 的问题
ArrayList<List<Integer>> temp = kSum(nums, target - nums[i], k - 1, i + 1, len);
if(temp != null) {
// 聚合 k - 1 Sum 的问题的结果
for (List<Integer> t : temp) {
t.add(0, nums[i]);
}
res.addAll(temp);
}
// 避免重复数字
while (i < len - k + 1 && nums[i] == nums[i + 1]) {
i++;
}
}
}
return res;
}
}
LeetCode题目：454. 4Sum II
Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.
To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.
Example:
Input:
A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]
Output:
2
Explanation:
The two tuples are:
1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0
以下代码的时间复杂度为 O(n^2)：
class Solution {
public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
int result = 0;
int N = A.length;
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int i = 0; i < N; i++) {
for(int j = 0; j < N; j++) {
int sum = A[i] + B[j];
map.put(sum, map.getOrDefault(sum, 0) + 1);
}
}
for(int i = 0; i < N; i++) {
for(int j = 0; j < N; j++) {
int sum = C[i] + D[j];
result = result + map.getOrDefault(-sum, 0);
}
}
return result;
}
}
