## 1. Two Sum
Given an array of integers nums
and an integer target
, return indices of the two numbers such that they add up to target
.
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]
Output: 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 <= 103
-109 <= nums[i] <= 109
-109 <= target <= 109
The brute force approach is simple. Loop through each element x and find if there is another value that equals to target - x
public int[] twoSum(int[] nums, int target){
for(int i=0; i<nums.length;i++){
for (int j=i+1; j<nums.length; j++){
if (nums[j]= target - nums[i]){
return new int[]{i,j};
}
}
}
throw new IllegalArgumentException("NO two sum solution");
}
Complexity Analysis
Time complexity : O(n^2). For each element, we try to find its complement by looping through the rest of array which takes O(n) time. Therefore, the time complexity is O(n^2)
Space complexity : O(1)
To improve our run time complexity, we need a more efficient way to check if the complement exists in the array. If the complement exists, we need to look up its index. What is the best way to maintain a mapping of each element in the array to its index? A hash table.
We reduce the look up time from O(n) to O(1) by trading space for speed. A hash table is built exactly for this purpose, it supports fast look up in near constant time. I say "near" because if a collision occurred, a look up could degenerate to O(n) time. But look up in hash table should be amortized O(1) time as long as the hash function was chosen carefully.
A simple implementation uses two iterations. In the first iteration, we add each element's value and its index to the table. Then, in the second iteration we check if each element's complement (target - nums[i] )exists in the table. Beware that the complement must not be nums[i] itself!
public int[] towSum(int[] nums, int target){
Map<Integer, Integer> map = new HashMap<>();
for (int i=0; i<nums.length; i++){
map.put(nums[i],i);
}
for (int i =0;i<nums.length; i++){
int complement = target - nums[i];
for(map.containsKey(complement) && map.get(complement)!=i){
return new int[]{i, map.get(complement)};
}
}
throw new IllegalArgumentException("NO two sum solution");
}
Complexity Analysis:
Time complexity : O(n) We traverse the list containing n elements exactly twice. Since the hash table reduces the look up time to O(1) , the time complexity is O(n) .
Space complexity : O(n) . The extra space required depends on the number of items stored in the hash table, which stores exactly n elements.
It turns out we can do it in one-pass. While we iterate and inserting elements into the table, we also look back to check if current element's complement already exists in the table. If it exists, we have found a solution and return immediately.
public int[] twoSum(int[] nums, int target){
Map<Integer, Integer> map = new HashMap<>{};
for(int i=0; i<nums.length; i++){
int complement = target - nums[i];
if(map.containsKey(complment)){
return new int[] {map.get(complement),i};
}
map.put(nums[i],i);
}
throw new IllegalArgumentException("No two sum solution");
}
vector<int> twoSum(vector<int> &numbers, int target){
//Key is the number and value is its index in the vector.
unordered_map<int, int> hash;
vector<int> result;
for(int i =0; i<numbers.size(); i++){
int numberToFind = target - numbers[i];
// if numberToFind is found in map, return them
if(hash.find(numberToFind)!=hash.end()){
result.push_back(hash[numberToFind]);
result.push_back(i);
return result;
}
//number was not found. Put it in the map.
hash[numbers[i]] = i;
}
return result;
}
Complexity Analysis:
class Solution:
def twoSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[int]
"""
h = {}
for i,num in enumerate(nums):
n = target - num
if n not in h:
h[num = i]
else:
return [h[n],i]
此处可能存在不合适展示的内容,页面不予展示。您可通过相关编辑功能自查并修改。
如您确认内容无涉及 不当用语 / 纯广告导流 / 暴力 / 低俗色情 / 侵权 / 盗版 / 虚假 / 无价值内容或违法国家有关法律法规的内容,可点击提交进行申诉,我们将尽快为您处理。