Two Sum：给出一个整数数组，返回两个数的下标值，令其和等于一个指定的目标值 #Leetcode

// Given nums = [2, 7, 11, 15], target = 9,

// Because nums[0] + nums[1] = 2 + 7 = 9,
// return [0, 1].


#include <iostream>
#include <vector>


int GetIndexByValue(const std::vector<int> &inArr, int value, int startIndex) {
    for (int i = startIndex; i < inArr.size(); ++i) {
        if (value == inArr[i]) {
            return i;
        }
    }

    return -1;
}

std::vector<int> GetIndicesOf2Addons(const std::vector<int> &inArr, int target) {
    for (int i = 0; i < inArr.size(); ++i) {
        int _2ndAddon = target - inArr[i];
        int indexOf2ndAddon = GetIndexByValue(inArr, _2ndAddon, i + 1);
        if (indexOf2ndAddon < 0) {
            std::cout << "Failed to find: " << _2ndAddon << "\n";
            continue;
        }

        return std::vector<int> {i, indexOf2ndAddon};
    }

    return std::vector<int>(0);
}


int main(int argc, char const *argv[]) {
    std::vector<int> nums {
        2, 3, 11, 15, -2
    };
    int target = 4;

    std::vector<int> result = GetIndicesOf2Addons(nums, target);

    if (result.empty()) {
        std::cout << "Failed.\n";
        return -1;
    }

    for (auto i: result) {
        std::cout << i << "\t";
    }
    std::cout << "\n";

    return 0;
}

// g++ two_sum.cpp -std=c++11 && ./a.out

 参考，https://cloud.tencent.com/developer/article/1010478，使用hash_map或者unordered_map实现：

// Given nums = [2, 7, 11, 15], target = 9,

// Because nums[0] + nums[1] = 2 + 7 = 9,
// return [0, 1].


#include <iostream>
#include <vector>
#include <unordered_map>


class Solution {
public:
    std::vector<int> twoSum(std::vector<int> &numbers, int target) {
        // Key is the number and value is its index in the std::vector.
        std::unordered_map<int, int> hash;

        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()) {
                return std::vector<int> {i, hash[numberToFind]};
            }

            // number was not found. Put it in the map.
            hash[numbers[i]] = i;
        }

        return std::vector<int>(0);
    }
};


int main(int argc, char const *argv[]) {
    std::vector<int> nums {
        2, 3, 11, 15
    };
    int target = 0;

    Solution sol;
    std::vector<int> res = sol.twoSum(nums, target);
    if (res.empty()) {
        std::cout << "Failed.\n";
        return -1;
    }

    for (auto i: res) {
        std::cout << i << "\t";
    }
    std::cout << "\n";

    return 0;
}

// g++ two_sum.cpp -std=c++11 && ./a.out

算法本身遍历一次，花费了 O(n) 的时间复杂度，遍历过程中的 find() 方法本身花费 O(log n)，所以该算法总时间复杂度为 O(nlog n)。