传送门:两数之和
解析:
1. 暴力遍历每一个数,使用unordered_map维护1 ~ (i - 1),区间中哪些数出现了
代码:
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> hash;
for (int i = 0, n = nums.size(); i < n; ++i) {
auto it = hash.find(target - nums[i]);
if (it != hash.end()) {
return {i, it->second};
}
hash[nums[i]] = i;
}
return {};
}
};