【题目】
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
【题意】
1. 从所给的列表中找出两个数，这两个数的和与给定的target值同样。返回这两个值的位置（保持相对顺序）
2. 所给測试用例保证有且仅仅有一个唯一解
【思路】
1. 绑定值和相应的位置 复杂度O(n)
2. 对值进行升序排序 复杂度O(nlogn)
3. 用两个指针p1,p2分别从前后两个方向逼近target。
当两指针之和小于target，则p1++
当两指针之和大于target。则p2--
【代码】
struct Node{
int index;
int value;
Node(){};
Node(int i, int v):index(i), value(v){};
};
bool compare(const Node &n1, const Node &n2){
return n1.value < n2.value;
}
class Solution {
public:
vector<int> twoSum(vector<int> &numbers, int target) {
vector<Node> nodes;
for(int i=0; i<numbers.size(); i++){
nodes.push_back(Node(i+1, numbers.at(i)));
}
sort(nodes.begin(), nodes.end(), compare);
int p1 = 0;
int p2 = nodes.size() - 1;
vector<int> indexs;
while(p1 < p2){
int sum = nodes.at(p1).value + nodes.at(p2).value;
if(sum == target){
indexs.push_back(min(nodes.at(p1).index, nodes.at(p2).index));
indexs.push_back(max(nodes.at(p1).index, nodes.at(p2).index));
break;
}
else{
if(sum < target) p1++;
else p2--;
}
}
return indexs;
}
};
本文转自mfrbuaa博客园博客，原文链接http://www.cnblogs.com/mfrbuaa/p/5212312.html，如需转载请自行联系原作者
