1 Two Sum 两数之和
Description:
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].
题目描述:
给定一个整数数组 nums 和一个目标值 target，请你在该数组中找出和为目标值的那两个整数，并返回他们的数组下标。
你可以假设每种输入只会对应一个答案。但是，你不能重复利用这个数组中同样的元素。
示例:
给定 nums = [2, 7, 11, 15], target = 9
因为 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]
思路:
- 暴力法: 对每个数, 遍历剩下的数求是否存在解. 时间复杂度O(n^2), 空间复杂度O(1)
- 一次遍历哈希法: 建立一个map存放相应元素对应的目标元素, 并在插入的时候检查, 时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
class Solution
{
public:
vector<int> twoSum(vector<int>& nums, int target)
{
vector<int> result;
// key为nums的值, value为nums下标
map<int, int> int_map;
for (int i = 0; i < nums.size(); i++)
{
// map.count()测试主键是否存在, 若存在返回1
if (int_map.count(nums[i]) != 0)
{
// push_back(elem)在容器最后位置添加一个元素elem
result.push_back(int_map[nums[i]]);
result.push_back(i);
break;
}
int_map[target- nums[i]] = i;
}
return result;
}
};
Java:
class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
int[] result = new int[2];
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (map.containsKey(complement)) {
result[0] = map.get(complement);
result[1] = i;
return result;
}
map.put(nums[i], i);
}
return result;
}
}
Python:
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
dic = {}
for index, num in enumerate(nums):
complement = target - num
if complement in dic:
return [dic[complement], index]
dic[num] = index
