[LeetCode] 1. Two Sum
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Given an array of integers nums
and an integer target
, return indices of the two numbers such that they add up totarget
.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
You can return the answer in any order.
Example 1:
**Input:** nums = [2,7,11,15], target = 9
**Output:** [0,1]
**Explanation:** Because nums[0] + nums[1] == 9, we return [0, 1].
Example 2:
**Input:** nums = [3,2,4], target = 6
**Output:** [1,2]
Example 3:
**Input:** nums = [3,3], target = 6
**Output:** [0,1]
Constraints:
2 <= nums.length <= 104
-109 <= nums[i] <= 109
-109 <= target <= 109
- Only one valid answer exists.
Follow-up: Can you come up with an algorithm that is less than O(n2)
time complexity?
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这道题给了我们一个数组,还有一个目标数target,让找到两个数字,使其和为 target,乍一看就感觉可以用暴力搜索,但是猜到 OJ 肯定不会允许用暴力搜索这么简单的方法,于是去试了一下,果然是 Time Limit Exceeded,这个算法的时间复杂度是 O(n^2)。那么只能想个 O(n) 的算法来实现,由于暴力搜索的方法是遍历所有的两个数字的组合,然后算其和,这样虽然节省了空间,但是时间复杂度高。一般来说,为了提高时间的复杂度,需要用空间来换,这算是一个 trade off 吧,但这里只想用线性的时间复杂度来解决问题,就是说只能遍历一个数字,那么另一个数字呢,可以事先将其存储起来,使用一个 HashMap,来建立数字和其坐标位置之间的映射,由于 HashMap 是常数级的查找效率,这样在遍历数组的时候,用 target 减去遍历到的数字,就是另一个需要的数字了,直接在 HashMap 中查找其是否存在即可,注意要判断查找到的数字不是第一个数字,比如 target 是4,遍历到了一个2,那么另外一个2不能是之前那个2,整个实现步骤为:先遍历一遍数组,建立 HashMap 映射,然后再遍历一遍,开始查找,找到则记录 index。代码如下:
C++ 解法一:
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> m;
vector<int> res;
for (int i = 0; i < nums.size(); ++i) {
m[nums[i]] = i;
}
for (int i = 0; i < nums.size(); ++i) {
int t = target - nums[i];
if (m.count(t) && m[t] != i) {
res.push_back(i);
res.push_back(m[t]);
break;
}
}
return res;
}
};
Java 解法一:
public class Solution {
public int[] twoSum(int[] nums, int target) {
HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();
int[] res = new int[2];
for (int i = 0; i < nums.length; ++i) {
m.put(nums[i], i);
}
for (int i = 0; i < nums.length; ++i) {
int t = target - nums[i];
if (m.containsKey(t) && m.get(t) != i) {
res[0] = i;
res[1] = m.get(t);
break;
}
}
return res;
}
}
或者可以写的更加简洁一些,把两个 for 循环合并成一个:
C++ 解法二:
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> m;
for (int i = 0; i < nums.size(); ++i) {
if (m.count(target - nums[i])) {
return {i, m[target - nums[i]]};
}
m[nums[i]] = i;
}
return {};
}
};
Java 解法二:
public class Solution {
public int[] twoSum(int[] nums, int target) {
HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();
int[] res = new int[2];
for (int i = 0; i < nums.length; ++i) {
if (m.containsKey(target - nums[i])) {
res[0] = i;
res[1] = m.get(target - nums[i]);
break;
}
m.put(nums[i], i);
}
return res;
}
}
Github 同步地址:
类似题目:
Two Sum III - Data structure design
Two Sum II - Input array is sorted
参考资料:
https://leetcode.com/problems/two-sum/
https://leetcode.com/problems/two-sum/discuss/3/Accepted-Java-O(n)-Solution
https://leetcode.com/problems/two-sum/discuss/13/Accepted-C++-O(n)-Solution
LeetCode All in One 题目讲解汇总(持续更新中...)
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解法二 输出为逆序 有误
Java的解法二输出结果错了
Solution in ReasonML/OCaml.
module LeetCode = {
let logl = l => l |> Array.of_list |> Js.log;
/* concatMap, drop, take and compose */
let rec range = (i: int, j: int) =>
if (i >= j) {
[];
} else {
[i, ...range(i + 1, j)];
};
let concatmap = (xs: list('a), f) => {
List.concat(List.map(x => f(x), xs));
};
let rec drop = (~n: int, xs: list('a)) =>
switch (xs) {
| [] => []
| l when n <= 0 => l
| [_, ...tail] => drop(~n=n - 1, tail)
};
let rec take = (~n: int, xs: list('a)) =>
switch (xs) {
| [] => []
| l when n <= 0 => l
| [x, ...tail] => n == 0 ? [] : [x, ...take(n - 1, tail)]
};
let slice = (list: list('a), start: int, range: int) => {
take(range, drop(start, list));
};
let compose = (f, g, x) => f(g(x));
let square = x => x * x;
let fact = x => x;
let square_o_fact = compose(square, fact);
module TwoSum = {
/* let odds = n => [0, ...range(0, n)]; */
/* let a = [? 2 * x | x <- 0 -- max_int ; x * x > 3]; */
let twoSum = (n, xs) => {
let ixs = Belt.List.zip([0, ...range(1, List.length(xs))], xs);
concatmap(ixs, ((i, x)) =>
concatmap(drop(i, ixs), ((j, y)) => x + y == n ? [[i, j]] : [])
);
};
let run = () => {
let res = twoSum(21, [0, 2, 11, 19, 90, 10]);
res |> logl;
/* List.iter(((i, j)) => Printf.printf("# Found: %d %d\n", i, j), res); */
};
};
let run = () => {
TwoSum.run();
};
};
LeetCode.run();
public static int[] dub(int[] nums,int target){
Map<Integer, Integer> map=new HashMap<Integer,Integer>();
int[] res = new int[2];
for (int i=0;i<nums.length;i++){
if (i == 0){
map.put(nums[0],0);
}
else {
int q=target-nums[i];
if (map.containsKey(q)){
res[0]=map.get(q);
res[1]=i;
return res;
}
map.put(nums[i],i);
}
}
return res;
}
public int[] twoSum2(int[] nums, int target){
HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();
int[] res = new int[2];
for (int i =0;i< nums.length;++i){
m.put(nums[i],i);
if (m.containsKey(target-nums[i])){
res[0]=m.get(target-nums[i]);
res[1]=i;
return res;
}
}return res;
}
rust解法
fn two_sum(nums: Vec<i32>, target: i32) -> Vec<i32> {
let mut m = HashMap::new();
for (i, j) in nums.iter().enumerate() {
if m.contains_key(&(target - j)) {
let mut res = vec![i as i32, *m.get(&(target - j)).unwrap()];
res.reverse();
return res;
}
m.insert(j, i as i32);
}
vec![]
}
go解法
func TwoSum(nums []int, target int) []int {
var res []int
resMap := make(map[int]int)
for k, v := range nums {
resMap[target-v] = k
}
for k, v := range nums {
if findK, isOk := resMap[v]; isOk && findK != k {
res = append(res, []int{k, findK}...)
break
}
}
return res
}
JS 解法
// Time complexity : O(n)
// Space complexity : O(n)
var twoSum = function (arr, target) {
if (arr.length < 2) return [];
let map = new Map();
for (let i = 0; i < arr.length; i++) {
let key = target - arr[i];
if (map.has(key)) {
return [map.get(key), i];
} else {
map.set(arr[i], i)
}
}
return [];
};