LeetCode-1-Two Sum
2015-04-23 20:50
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Given an array of integers, 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.
Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2
主要问题是要返回元素的原始下标,如果只是返回元素本身,则可以直接将数组排序,然后从两边向中间遍历,寻找目标数;为了输出下标,建立一个map来存储。
代码如下:
class Solution {
public:
vector<int> twoSum(vector<int> &numbers, int target) {
vector<int > result;
map<int,int> temp;
int len=numbers.size();
for(int i=0;i<len;i++){
int gap=target-numbers[i];
if(temp.find(gap)!=temp.end()){
result.push_back(temp[gap]);
result.push_back(i+1);
return result;
}
temp[numbers[i]]=i+1;
}
}
};
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.
Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2
主要问题是要返回元素的原始下标,如果只是返回元素本身,则可以直接将数组排序,然后从两边向中间遍历,寻找目标数;为了输出下标,建立一个map来存储。
代码如下:
class Solution {
public:
vector<int> twoSum(vector<int> &numbers, int target) {
vector<int > result;
map<int,int> temp;
int len=numbers.size();
for(int i=0;i<len;i++){
int gap=target-numbers[i];
if(temp.find(gap)!=temp.end()){
result.push_back(temp[gap]);
result.push_back(i+1);
return result;
}
temp[numbers[i]]=i+1;
}
}
};
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