1. Two Sum【easy】
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].
解法一：
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> u_map;
vector<int> result;
for (int i = ; i < nums.size(); ++i)
{
if (u_map.find(target - nums[i]) != u_map.end())
{
result.push_back(u_map[target - nums[i]]);
result.push_back(i);
return result;
}
u_map.insert(make_pair(nums[i], i));
}
return result;
}
};
map搞一把
解法二：
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> map;
int n = (int)nums.size();
for (int i = ; i < n; i++) {
auto p = map.find(target - nums[i]);
if (p != map.end()) {
return {p->second, i};
}
map[nums[i]] = i;
}
}
};
比较简洁的写法
扩展见：