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].
1、直接循环
/**
* Question Link: https://leetcode.com/problems/two-sum/
* Primary idea: Loop through each element x and find if there is another value that equals to target - x.
*
* Time Complexity: O(n^2), Space Complexity: O(1)
*/
func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
for i in 0..<nums.count {
for j in 0..<nums.count {
if nums[j] == target - nums[i] {
return [i ,j]
}
}
}
fatalError("No valid outputs")
}
2、对数组迭代时用哈希表来记录迭代过的数据，每次迭代都可通过哈希表快速查找结果
/**
* Question Link: https://leetcode.com/problems/two-sum/
* Primary idea: Traverse the array and store target - nums[i] in a dict
*
* Time Complexity: O(n), Space Complexity: O(n)
*/
func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
//nums的值为key，索引为value
var dict = [Int: Int]()
for (i, num) in nums.enumerated() {
//每次迭代，通过dict[target - num]快速查询
if let lastIndex = dict[target - num] {
//如果有结果，则表示nums[i] + nums[lastIndex] = target
return [lastIndex, i]
}
dict[num] = i
}
fatalError("No valid outputs")
}
