万事开头难，算是开始刷leetcode了吧，不知道能不能坚持下去！在这放下一句，不坚持是小狗！——jc,15.10.25

第一题能就是一个纯属熟悉环境，从网上找了python版答案走了一遍流程：

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

python

class Solution:
    # @return a tuple, (index1, index2)
    def twoSum(self, num, target):
        dictMap = {}
        for index, value in enumerate(num):
            if target - value in dictMap:
                return dictMap[target - value] + 1, index + 1
            dictMap[value] = index

然后看到了题解：

O( n^2 ) runtime, O(1) space – Brute force:

The brute force approach is simple. Loop through each element x and find if there is another value that equals to target – x. As finding another value requires looping through the rest of array, its runtime complexity is O( n^2 ).

O(n) runtime, O(n) space – Hash table:

We could reduce the runtime complexity of looking up a value to O(1) using a hash map that maps a value to its index.

题解大意就是用直接遍历(暴力破解)的话就是O( n^2 ),但是用哈希表就是O(n)的复杂度。具体到python语言就可以使用内置的字典(dict)直接编写了。

后来看了看C++的题解，尝试写了一个C++版本的代码：

C++(数组)

#define MAX 200002
#define OFFSET 100000

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> answer;
        int num_id[MAX];
        memset(num_id, 0, sizeof(num_id));
        int length = nums.size();
        int rest;
        for (int i = 0; i < length; i++){
            rest = target - nums[i];
            if(num_id[rest + OFFSET]){
            //由于rest和nums都可能为负数，数组OFFSET以前表示负数
                answer.push_back(num_id[rest + OFFSET]);
                answer.push_back(i + 1);
                return answer;
            }else{
                num_id[nums[i] + OFFSET] = i + 1;
            }
        }
    }
};

C++(STL map)

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> answer;
        map<int ,int> num_index;//key:num value:index
        int length = nums.size();
        int rest;
        for (int i = 0; i < length; i++){
            rest = target - nums[i];
            if(num_index[rest]){
                answer.push_back(num_index[rest]);
                answer.push_back(i + 1);
                return answer;
            }else{
                num_index[nums[i]] = i + 1;
            }
        }
    }
};

Java

(题解答案，暂时不会java)

public int[] twoSum(int[] numbers, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < numbers.length; i++) {
        int x = numbers[i];
        if (map.containsKey(target - x)) {
            return new int[] { map.get(target - x) + 1, i + 1 };
        }
        map.put(x, i);
    }
    throw new IllegalArgumentException("No two sum solution");
}

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• 参考：

[1] http://blog.csdn.net/hcbbt/article/details/43966403 [LeetCode] 001. Two Sum (Medium) (C++/Java/Python)