LeetCode:1. Add Two Numbers
2017-05-11 00:13
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题目:
LeetCode:1. Add Two Numbers
描述：
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.
样例：
分析：
思路如下：
1) 利用hashmap来存储数组元素以及其对应索引值，提高检索效率
2) 通过nTarget - vecNum[i]来获取对应的检索目标元素
3) 返回满足条件的索引值
代码：
备注：
LC上大神的 0ms 的代码：
LeetCode:1. Add Two Numbers
描述：
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.
样例：
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
分析：
思路如下：
1) 利用hashmap来存储数组元素以及其对应索引值，提高检索效率
2) 通过nTarget - vecNum[i]来获取对应的检索目标元素
3) 返回满足条件的索引值
代码：
vector<int> twoSum(vector<int>& vecNum, int nTarget) {
vector<int> vecTemp;
unordered_map<int, int> hashMapTemp;
for (int i = 0; i < vecNum.size(); i++)
{
hashMapTemp[vecNum[i]] = i;
}
for (int i = 0; i < vecNum.size(); i++)
{
const int nPart = nTarget - vecNum[i];
if (hashMapTemp.find(nPart) != hashMapTemp.end() && hashMapTemp[nPart] > i)
{
vecTemp.push_back(i);
vecTemp.push_back(hashMapTemp[nPart]);
break;
}
}
return vecTemp;
}备注：
LC上大神的 0ms 的代码：
vector<int> twoSum(vector<int> &numbers, int target)
{
//Key is the number and value is its index in the vector.
unordered_map<int, int> hash;
vector<int> result;
for (int i = 0; i < numbers.size(); i++) {
int numberToFind = target - numbers[i];
//if numberToFind is found in map, return them
if (hash.find(numberToFind) != hash.end()) {
//+1 because indices are NOT zero based
result.push_back(hash[numberToFind] + 1);
result.push_back(i + 1);
return result;
}
//number was not found. Put it in the map.
hash[numbers[i]] = i;
}
return result;
}
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