---
comments: true
---
# 10.4 Hash optimization strategies
In algorithm problems, **we often reduce the time complexity of algorithms by replacing linear search with hash search**. Let's use an algorithm problem to deepen understanding.
!!! question
Given an integer array `nums` and a target element `target`, please search for two elements in the array whose "sum" equals `target`, and return their array indices. Any solution is acceptable.
## 10.4.1 Linear search: trading time for space
Consider traversing all possible combinations directly. As shown in the Figure 10-9 , we initiate a two-layer loop, and in each round, we determine whether the sum of the two integers equals `target`. If so, we return their indices.
{ class="animation-figure" }
Figure 10-9 Linear search solution for two-sum problem
The code is shown below:
=== "Python"
```python title="two_sum.py"
def two_sum_brute_force(nums: list[int], target: int) -> list[int]:
"""方法一：暴力枚举"""
# 两层循环，时间复杂度为 O(n^2)
for i in range(len(nums) - 1):
for j in range(i + 1, len(nums)):
if nums[i] + nums[j] == target:
return [i, j]
return []
```
=== "C++"
```cpp title="two_sum.cpp"
/* 方法一：暴力枚举 */
vector twoSumBruteForce(vector &nums, int target) {
int size = nums.size();
// 两层循环，时间复杂度为 O(n^2)
for (int i = 0; i < size - 1; i++) {
for (int j = i + 1; j < size; j++) {
if (nums[i] + nums[j] == target)
return {i, j};
}
}
return {};
}
```
=== "Java"
```java title="two_sum.java"
/* 方法一：暴力枚举 */
int[] twoSumBruteForce(int[] nums, int target) {
int size = nums.length;
// 两层循环，时间复杂度为 O(n^2)
for (int i = 0; i < size - 1; i++) {
for (int j = i + 1; j < size; j++) {
if (nums[i] + nums[j] == target)
return new int[] { i, j };
}
}
return new int[0];
}
```
=== "C#"
```csharp title="two_sum.cs"
/* 方法一：暴力枚举 */
int[] TwoSumBruteForce(int[] nums, int target) {
int size = nums.Length;
// 两层循环，时间复杂度为 O(n^2)
for (int i = 0; i < size - 1; i++) {
for (int j = i + 1; j < size; j++) {
if (nums[i] + nums[j] == target)
return [i, j];
}
}
return [];
}
```
=== "Go"
```go title="two_sum.go"
/* 方法一：暴力枚举 */
func twoSumBruteForce(nums []int, target int) []int {
size := len(nums)
// 两层循环，时间复杂度为 O(n^2)
for i := 0; i < size-1; i++ {
for j := i + 1; j < size; j++ {
if nums[i]+nums[j] == target {
return []int{i, j}
}
}
}
return nil
}
```
=== "Swift"
```swift title="two_sum.swift"
/* 方法一：暴力枚举 */
func twoSumBruteForce(nums: [Int], target: Int) -> [Int] {
// 两层循环，时间复杂度为 O(n^2)
for i in nums.indices.dropLast() {
for j in nums.indices.dropFirst(i + 1) {
if nums[i] + nums[j] == target {
return [i, j]
}
}
}
return [0]
}
```
=== "JS"
```javascript title="two_sum.js"
/* 方法一：暴力枚举 */
function twoSumBruteForce(nums, target) {
const n = nums.length;
// 两层循环，时间复杂度为 O(n^2)
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
if (nums[i] + nums[j] === target) {
return [i, j];
}
}
}
return [];
}
```
=== "TS"
```typescript title="two_sum.ts"
/* 方法一：暴力枚举 */
function twoSumBruteForce(nums: number[], target: number): number[] {
const n = nums.length;
// 两层循环，时间复杂度为 O(n^2)
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
if (nums[i] + nums[j] === target) {
return [i, j];
}
}
}
return [];
}
```
=== "Dart"
```dart title="two_sum.dart"
/* 方法一： 暴力枚举 */
List twoSumBruteForce(List nums, int target) {
int size = nums.length;
// 两层循环，时间复杂度为 O(n^2)
for (var i = 0; i < size - 1; i++) {
for (var j = i + 1; j < size; j++) {
if (nums[i] + nums[j] == target) return [i, j];
}
}
return [0];
}
```
=== "Rust"
```rust title="two_sum.rs"
/* 方法一：暴力枚举 */
pub fn two_sum_brute_force(nums: &Vec, target: i32) -> Option> {
let size = nums.len();
// 两层循环，时间复杂度为 O(n^2)
for i in 0..size - 1 {
for j in i + 1..size {
if nums[i] + nums[j] == target {
return Some(vec![i as i32, j as i32]);
}
}
}
None
}
```
=== "C"
```c title="two_sum.c"
/* 方法一：暴力枚举 */
int *twoSumBruteForce(int *nums, int numsSize, int target, int *returnSize) {
for (int i = 0; i < numsSize; ++i) {
for (int j = i + 1; j < numsSize; ++j) {
if (nums[i] + nums[j] == target) {
int *res = malloc(sizeof(int) * 2);
res[0] = i, res[1] = j;
*returnSize = 2;
return res;
}
}
}
*returnSize = 0;
return NULL;
}
```
=== "Kotlin"
```kotlin title="two_sum.kt"
/* 方法一：暴力枚举 */
fun twoSumBruteForce(nums: IntArray, target: Int): IntArray {
val size = nums.size
// 两层循环，时间复杂度为 O(n^2)
for (i in 0..
This method has a time complexity of $O(n^2)$ and a space complexity of $O(1)$, which is very time-consuming with large data volumes.
## 10.4.2 Hash search: trading space for time
Consider using a hash table, with key-value pairs being the array elements and their indices, respectively. Loop through the array, performing the steps shown in the figures below each round.
1. Check if the number `target - nums[i]` is in the hash table. If so, directly return the indices of these two elements.
2. Add the key-value pair `nums[i]` and index `i` to the hash table.
=== "<1>"
{ class="animation-figure" }
=== "<2>"
{ class="animation-figure" }
=== "<3>"
{ class="animation-figure" }
Figure 10-10 Help hash table solve two-sum
The implementation code is shown below, requiring only a single loop:
=== "Python"
```python title="two_sum.py"
def two_sum_hash_table(nums: list[int], target: int) -> list[int]:
"""方法二：辅助哈希表"""
# 辅助哈希表，空间复杂度为 O(n)
dic = {}
# 单层循环，时间复杂度为 O(n)
for i in range(len(nums)):
if target - nums[i] in dic:
return [dic[target - nums[i]], i]
dic[nums[i]] = i
return []
```
=== "C++"
```cpp title="two_sum.cpp"
/* 方法二：辅助哈希表 */
vector twoSumHashTable(vector &nums, int target) {
int size = nums.size();
// 辅助哈希表，空间复杂度为 O(n)
unordered_map dic;
// 单层循环，时间复杂度为 O(n)
for (int i = 0; i < size; i++) {
if (dic.find(target - nums[i]) != dic.end()) {
return {dic[target - nums[i]], i};
}
dic.emplace(nums[i], i);
}
return {};
}
```
=== "Java"
```java title="two_sum.java"
/* 方法二：辅助哈希表 */
int[] twoSumHashTable(int[] nums, int target) {
int size = nums.length;
// 辅助哈希表，空间复杂度为 O(n)
Map dic = new HashMap<>();
// 单层循环，时间复杂度为 O(n)
for (int i = 0; i < size; i++) {
if (dic.containsKey(target - nums[i])) {
return new int[] { dic.get(target - nums[i]), i };
}
dic.put(nums[i], i);
}
return new int[0];
}
```
=== "C#"
```csharp title="two_sum.cs"
/* 方法二：辅助哈希表 */
int[] TwoSumHashTable(int[] nums, int target) {
int size = nums.Length;
// 辅助哈希表，空间复杂度为 O(n)
Dictionary dic = [];
// 单层循环，时间复杂度为 O(n)
for (int i = 0; i < size; i++) {
if (dic.ContainsKey(target - nums[i])) {
return [dic[target - nums[i]], i];
}
dic.Add(nums[i], i);
}
return [];
}
```
=== "Go"
```go title="two_sum.go"
/* 方法二：辅助哈希表 */
func twoSumHashTable(nums []int, target int) []int {
// 辅助哈希表，空间复杂度为 O(n)
hashTable := map[int]int{}
// 单层循环，时间复杂度为 O(n)
for idx, val := range nums {
if preIdx, ok := hashTable[target-val]; ok {
return []int{preIdx, idx}
}
hashTable[val] = idx
}
return nil
}
```
=== "Swift"
```swift title="two_sum.swift"
/* 方法二：辅助哈希表 */
func twoSumHashTable(nums: [Int], target: Int) -> [Int] {
// 辅助哈希表，空间复杂度为 O(n)
var dic: [Int: Int] = [:]
// 单层循环，时间复杂度为 O(n)
for i in nums.indices {
if let j = dic[target - nums[i]] {
return [j, i]
}
dic[nums[i]] = i
}
return [0]
}
```
=== "JS"
```javascript title="two_sum.js"
/* 方法二：辅助哈希表 */
function twoSumHashTable(nums, target) {
// 辅助哈希表，空间复杂度为 O(n)
let m = {};
// 单层循环，时间复杂度为 O(n)
for (let i = 0; i < nums.length; i++) {
if (m[target - nums[i]] !== undefined) {
return [m[target - nums[i]], i];
} else {
m[nums[i]] = i;
}
}
return [];
}
```
=== "TS"
```typescript title="two_sum.ts"
/* 方法二：辅助哈希表 */
function twoSumHashTable(nums: number[], target: number): number[] {
// 辅助哈希表，空间复杂度为 O(n)
let m: Map = new Map();
// 单层循环，时间复杂度为 O(n)
for (let i = 0; i < nums.length; i++) {
let index = m.get(target - nums[i]);
if (index !== undefined) {
return [index, i];
} else {
m.set(nums[i], i);
}
}
return [];
}
```
=== "Dart"
```dart title="two_sum.dart"
/* 方法二： 辅助哈希表 */
List twoSumHashTable(List nums, int target) {
int size = nums.length;
// 辅助哈希表，空间复杂度为 O(n)
Map dic = HashMap();
// 单层循环，时间复杂度为 O(n)
for (var i = 0; i < size; i++) {
if (dic.containsKey(target - nums[i])) {
return [dic[target - nums[i]]!, i];
}
dic.putIfAbsent(nums[i], () => i);
}
return [0];
}
```
=== "Rust"
```rust title="two_sum.rs"
/* 方法二：辅助哈希表 */
pub fn two_sum_hash_table(nums: &Vec, target: i32) -> Option> {
// 辅助哈希表，空间复杂度为 O(n)
let mut dic = HashMap::new();
// 单层循环，时间复杂度为 O(n)
for (i, num) in nums.iter().enumerate() {
match dic.get(&(target - num)) {
Some(v) => return Some(vec![*v as i32, i as i32]),
None => dic.insert(num, i as i32),
};
}
None
}
```
=== "C"
```c title="two_sum.c"
/* 哈希表 */
typedef struct {
int key;
int val;
UT_hash_handle hh; // 基于 uthash.h 实现
} HashTable;
/* 哈希表查询 */
HashTable *find(HashTable *h, int key) {
HashTable *tmp;
HASH_FIND_INT(h, &key, tmp);
return tmp;
}
/* 哈希表元素插入 */
void insert(HashTable *h, int key, int val) {
HashTable *t = find(h, key);
if (t == NULL) {
HashTable *tmp = malloc(sizeof(HashTable));
tmp->key = key, tmp->val = val;
HASH_ADD_INT(h, key, tmp);
} else {
t->val = val;
}
}
/* 方法二：辅助哈希表 */
int *twoSumHashTable(int *nums, int numsSize, int target, int *returnSize) {
HashTable *hashtable = NULL;
for (int i = 0; i < numsSize; i++) {
HashTable *t = find(hashtable, target - nums[i]);
if (t != NULL) {
int *res = malloc(sizeof(int) * 2);
res[0] = t->val, res[1] = i;
*returnSize = 2;
return res;
}
insert(hashtable, nums[i], i);
}
*returnSize = 0;
return NULL;
}
```
=== "Kotlin"
```kotlin title="two_sum.kt"
/* 方法二：辅助哈希表 */
fun twoSumHashTable(nums: IntArray, target: Int): IntArray {
val size = nums.size
// 辅助哈希表，空间复杂度为 O(n)
val dic = HashMap()
// 单层循环，时间复杂度为 O(n)
for (i in 0..
This method reduces the time complexity from $O(n^2)$ to $O(n)$ by using hash search, greatly improving the running efficiency.
As it requires maintaining an additional hash table, the space complexity is $O(n)$. **Nevertheless, this method has a more balanced time-space efficiency overall, making it the optimal solution for this problem**.