原题

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].

题解

暴力解法

暴力解法使用两层for循环,遍历每一种存在的情况。

[code lang=”cpp”]
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> a;
for (int i = 0; i < nums.size(); ++i) {
for (int j = i + 1; j < nums.size(); ++j) {
if (nums[i] + nums[j] == target) {
a.push_back(i);
a.push_back(j);
return a;
}
}
}
return a;
}
};
[/code]

时间复杂度:O(n^2)

空间复杂度:O(1)

哈希table

使用空间换时间,在遍历nums时,将nums的数据,键值对颠倒重写一份到map中。同时,寻找满足题解(map.find(target 1 nums[i]) != map.end())的答案。

[code lang=”cpp”]
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
map<int, int> hp;
vector<int> a;
for (int i = 0; i < nums.size(); ++i) {
if (hp.find(target – nums[i]) != hp.end()) {
a.push_back(i);
a.push_back(hp[target – nums[i]]);
return a;
}
hp[nums[i]] = i;
}
return a;
}
};
[/code]

时间复杂度:O(n)

空间复杂度:O(n)


文章来源:胡小旭 => [LeetCode]Two Sum

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