Leetcode 1. Two Sum
2017-12-28 19:24
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题目的意思是找出一个数组中和为给定的目标数的两个数。
方法1:最简单的,两重for循环,遍历数组。复杂度O(n2).
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int>output;
for(int i=0; i<nums.size(); ++i)
{
for(int j=i+1; j<nums.size(); ++j)
{
if(nums[i]+nums[j]==target)
{
output.push_back(i);
output.push_back(j);
return output;
}
}
}
}
};
方法2:用哈希表,key存数组元素,value存对应的下标值,遍历数组,当查找到target-nums[i]时,此时元素的下标和target-nums[i]的value值就是所求结果。时间复杂度为O (n)的方法,用了额外空间。class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
map<int, int>m;
for(int i=0; i<nums.size(); ++i)
{
if(m.count(target-nums[i]))
return vector<int>{min(i, m[target-nums[i]]), max(i, m[target-nums[i]])};
else
m[nums[i]]=i;
}
}
};
方法1:最简单的,两重for循环,遍历数组。复杂度O(n2).
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int>output;
for(int i=0; i<nums.size(); ++i)
{
for(int j=i+1; j<nums.size(); ++j)
{
if(nums[i]+nums[j]==target)
{
output.push_back(i);
output.push_back(j);
return output;
}
}
}
}
};
方法2:用哈希表,key存数组元素,value存对应的下标值,遍历数组,当查找到target-nums[i]时,此时元素的下标和target-nums[i]的value值就是所求结果。时间复杂度为O (n)的方法,用了额外空间。class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
map<int, int>m;
for(int i=0; i<nums.size(); ++i)
{
if(m.count(target-nums[i]))
return vector<int>{min(i, m[target-nums[i]]), max(i, m[target-nums[i]])};
else
m[nums[i]]=i;
}
}
};
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