最简单高效的解法,就是用哈希表

其他解法
解法一:对于每个元素nums[i],遍历右侧元素是否存在target-nums[i]。时间复杂度o(n^2)
解法二:先对数组排序,然后双指针分别指向数组开头和结尾,两元素相加,若和>target,则右指针左移,若和<target,则左指针右移,直至找到答案。时间复杂度取决于排序方法

zhuhui 2024年06月04日 打卡

Additionally to hash-table you can use approach with 2 pointers.
Formally time complexity will be the same O(n), but technically it should be 2 times faster.
But unfortunately it also comes with code duplication.. 🤷🏻

function twoSum(nums: number[], target: number): number[] {
 const map: Record<number, number> = {}
 const n = nums.length
 for (let l = 0, r = n - 1; l <= r; l++, r--) {
 let v = nums[l]
 let diff = target - v
 if (map[diff] !== undefined) return [map[diff], l]
 map[v] = l
 v = nums[r]
 diff = target - v
 if (map[diff] !== undefined) return [map[diff], r]
 map[v] = r
 }
 return []
}

我直接return num1 + num2,然后你告诉我还能这么些

小白请教一下,在java的实现中为什么for的循环条件可以设置为空,这样不会导致无限循环吗