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].思路
- 直接一个排序，O(nlgn)，然后一个查找，O(n)
- 使用unordered_map<int,int>可直接实现O(n)的解法
代码
- O(nlgn)的解法
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> res;
int len = nums.size();
if(len == 0) return res;
vector<pair<int, int>> vec;
for(int i = 0; i < len; ++i){
vec.push_back(pair<int, int>(nums[i], i));
}
sort(vec.begin(), vec.end(),
[](const pair<int, int> &a, const pair<int, int>&b){ return a.first < b.first;});
int i = 0, j = len - 1, tmp;
while(i < j){
tmp = vec[i].first + vec[j].first;
if(tmp == target){
res.push_back(min(vec[i].second, vec[j].second));
res.push_back(max(vec[i].second, vec[j].second));
break;
}
else if(tmp < target)
i++;
else j--;
}
return res;
}
};- O(n)的解法
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> res;
int len = nums.size();
if(len == 0) return res;
unordered_map<int, int> hash;
int needToFind = 0;
for(int i = 0; i < len; ++i){
needToFind = target - nums[i];
if(hash.find(needToFind) != hash.end()){
res.push_back(min(i, hash[needToFind]));
res.push_back(max(i, hash[needToFind]));
break;
}
else{
hash.insert(pair<int, int>(nums[i], i));
}
}
return res;
}
};内容来源于网络如有侵权请私信删除
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