leetcode two sum

很经典的一道题。一个数组里找两个数相加等于目标值，输出两个数的位置。

实在没想出来什么好办法，有一个思路，按照快排的想法，每次选取一个随机标记，分为左右，然后根据标记和目标值的大小比较，缩小搜索范围，最终时间复杂度是O(nlogn).

不过想想就觉得很难code出来，所以最终还是采用了排序再二分查找的办法，很普通的算法。

另外要注意两个相同的数也要处理。

class Solution {public:    vector<int> twoSum(vector<int> &numbers, int target) {        vector<int> res;    vector<int> indexes;    int i;    for (i = 0; i < numbers.size(); i++)    {    indexes.push_back(numbers[i]);    }    sort(numbers.begin(), numbers.end());    int n = numbers.size();    for (i = 0; i < n; i++){    int num = numbers[i];    int left = i+1, right = n-1, mid = (left + right) / 2 ;    int tmpsum = -1;    while (left <= right){    tmpsum = num + numbers[mid];    if (tmpsum == target)break;    else if (tmpsum < target)left = mid + 1;    else right = mid - 1;        mid = (left + right) / 2;    }    if (tmpsum == target){    int a = -1, b = -1;    for (int j = 0; j < n; j++){    if (indexes[j] == numbers[i] || indexes[j] == numbers[mid]){    if (a < 0)a = j + 1;    else{    b = j + 1; break;    }    }    }        res.push_back(a);    res.push_back(b);    break;    }    }        return res;    }};

另一种用map的解法我也试过，但是时间花得多的多。而且用的是unordered_map，不知道是不是编译器的原因。

下面是别人的类似代码，写得不太漂亮：

vector<int> twoSum(vector<int> &numbers, int target) {    int index1 = 0, index2 = 0;    // Special Case: target = a + a, cannot be solved by map    for (int i = 0; i < numbers.size(); ++i) {        if (numbers[i] == target / 2) {            if (index1 == 0) index1 = i + 1; else index2 = i + 1;        }    }    if (index1 > 0 && index2 > 0) {        vector<int> result { index1, index2 };        return result;    }    unordered_map<int, int> m;    for (int i = 0; i < numbers.size(); ++i) m[numbers[i]] = i + 1;    for (int i = 0; i < numbers.size(); ++i) {        if (m[target - numbers[i]] && m[target - numbers[i]] != i + 1) {            index1 = m[numbers[i]];            index2 = m[target - numbers[i]];            break;        }    }    if (index1 > index2) { index1 ^= index2; index2 ^= index1; index1 ^= index2;  }    vector<int> result { index1, index2 };    return result;}