LeetCode 1 — Two Sum（C++ Java Python）

题目：http://oj.leetcode.com/problems/two-sum/

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

题目翻译：

给定一个整数数组，找到两个数，使得它们的和等于一个特定的目标数。
函数twoSum应返回两个数的索引（index），使它们加起来等于目标，其中索引1必须小于索引2。注意返回的答案（包括索引1和索引2 ）不是从零开始的。
可以假设每个输入有且仅有一个答案。
输入：数字 = {2,7,11,15}，目标 = 9
输出：索引1 = 1，索引2 = 2

分析：

        首先复制一份array，对其进行排序，找到符合条件的两个数，再在原数组里找到index。

C++实现：

class Solution {
public:
    vector twoSum(vector &numbers, int target) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        vector num = numbers;
        std::sort(num.begin(), num.end());
        
        int length = numbers.size();
        int left = 0;
        int right = length - 1;
        int sum = 0;
        
        vector index;
        
        while(left < right)
        {
        	sum = num[left] + num[right];
        	
            if(sum == target)
            {
            	for(int i = 0; i < length; ++i)
            	{
            		if(numbers[i] == num[left])
            		{
            			index.push_back(i + 1);
            		}
            		else if(numbers[i] == num[right])
            		{
            			index.push_back(i + 1);
            		}
            		if(index.size() == 2)
            		{
            			break;
            		}
            	}
                break;
            }
            else if(sum > target)
            {
                --right;
            }
            else
            {
                ++left;
            }
        }
        
        return index;
    }
};

Java实现：

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int[] num = numbers.clone();
		Arrays.sort(num);

		int length = numbers.length;
		int left = 0;
		int right = length - 1;
		int sum = 0;

		ArrayList index = new ArrayList();

		while (left < right) {
			sum = num[left] + num[right];

			if (sum == target) {
				for (int i = 0; i < length; ++i) {
					if (numbers[i] == num[left]) {
						index.add(i + 1);
					} else if (numbers[i] == num[right]) {
						index.add(i + 1);
					}
					if (index.size() == 2) {
						break;
					}
				}
				break;
			} else if (sum > target) {
				--right;
			} else {
				++left;
			}
		}

		int[] result = new int[2];

		result[0] = index.get(0);
		result[1] = index.get(1);

		return result;
	}

	public static void main(String[] args) {
		Solution slt = new Solution();

		int[] numbers = { 2, 7, 11, 15 };
		int target = 9;

		int[] index = new int[2];

		index = slt.twoSum(numbers, target);

		System.out.println("index1=" + index[0] + ", index2=" + index[1]);
	}
}

Python实现：

class Solution:
    # @return a tuple, (index1, index2)
    def twoSum(self, num, target):
     
        numbers = sorted(num)

        length = len(num)
        left = 0
        right = length - 1

        index = []

        while left < right: 
            sums = numbers[left] + numbers[right]

            if sums == target:
                for i in range(length):
                    if num[i] == numbers[left]:
                        index.append(i + 1)
                    elif num[i] == numbers[right]:
                        index.append(i + 1)
                    
                    if len(index) == 2:
                        break

                break
            elif sums > target:
                right -= 1
            else:
                left += 1

        result = tuple(index)

        return result

        感谢阅读，欢迎评论！