Two Sum

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

分析

1.数字顺序并非有序。

2.index1 < index2 ，并且index1与index2起始于1。

3.存在唯一解。

方案

1.存储value和index对value进行排序，使用Quick Sort, 时间复杂度为O(nlogn)，从两端向中间逼近，复杂度为O(n),总复杂度为O(nlogn)。

2.Hash,C++中使用关联容器map，map为红黑树，查找效率为O(logn),另外需要便利数组中每个元素，时间复杂度为O(n)，故总时间复杂度为O(nlogn)。

结论

从以上分析来看，时间复杂度相同，使用排序后查找占用48ms，使用map占用96ms。

CODE

struct Node{
int value;
int index;
};

bool cmp(Node a, Node b) {
return a.value < b.value;
}

class Solution {
public:
vector<int> twoSum(vector<int> &numbers, int target) {
vector<Node> ary;
for (int i = 0; i < numbers.size(); ++i) {
Node t;
t.value = numbers[i];
t.index = i + 1;
ary.push_back(t);
}

sort(ary.begin(), ary.end(),cmp);

vector<int> ans;
for (int i = 0, j = ary.size()-1; i < j; ) {
int sum = ary[i].value + ary[j].value;
if (sum == target) {
int index1 = ary[i].index;
int index2 = ary[j].index;
if(index1 < index2) {
ans.push_back(index1);
ans.push_back(index2);
} else {
ans.push_back(index2);
ans.push_back(index1);
}
break;
} else if(sum < target) {
i++;
} else {
j--;
}
}
return ans;
}
};

class Solution {
public:
vector<int> twoSum(vector<int> &numbers, int target) {
map<int, int> mymap;
for (int i = 0; i < numbers.size(); ++i) {
mymap[numbers[i]] = i+1;
}

vector<int> result;
for (int i = 0; i < numbers.size(); ++i) {
int a = numbers[i];
int b = target - a;
map<int, int>::iterator it = mymap.begin();
it = mymap.find(b);
if (it != mymap.end()) {
// find it
int index1 = i+1;
int index2 = it->second;
if (index1 < index2) {
result.push_back(index1);
result.push_back(index2);
} else if(index1 > index2) {
result.push_back(index2);
result.push_back(index2);
} else {
continue;
}
break;
}
}
return result;
}
};