关关的刷题日记14——Leetcode 167. Two Sum II - Input array is sorted
关关的刷题日记14——Leetcode 167. Two Sum II - Input array is sorted
关小刷刷题14 – Leetcode 167. Two Sum II - Input array is sorted
题目
Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution and you may not use the same element twice.
Input: numbers={2, 7, 11, 15}, target=9Output: index1=1, index2=2
这题与two sum的区别是，数组是按照从小到大的顺序排好序的，参考一下刷题日记11中的两种做法。
方法1
方法1：双重指针，时间复杂度O(n)。对于排好序的数组这种方法非常好！
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
int i=0, j=numbers.size()-1;
vector<int>output(2,0);
while(true)
{
int sum=numbers[i]+numbers[j];
if(sum>target)
j--;
else if(sum<target)
i++;
else
{
output[0]=i+1;
output[1]=j+1;
break;
}
}
return output;
}
};方法2
方法2：哈希表，时间复杂度O(n).
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
vector<int>output(2,0);
hash_map<int, int>m;
for(int i=0; i<numbers.size(); i++)
{
if(m.find(target-x)!=m.end())
m[x]=i+1;
else
{
output[0]=min(m[x], m[target-x]);
output[1]=max(m[x], m[target-x]);
break;
}
}
return output;
}
};只有非常努力，才能看起来毫不费力，加油！
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