LeetCode-TwoSum

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作者：disappearedgod

文章出处：http://blog.csdn.net/disappearedgod/article/details/23634067

时间：2014-4-13

题目

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

思路：

1.克隆排序

2.双指针查找

3.对应返回位置

C++ 版本

class Solution {public:    vector<int> twoSum(vector<int> &numbers, int target) {        vector<int> num = numbers;        std::sort(num.begin(),num.end());                int length = num.size();        int left = 0;        int right = length -1;        int sum = 0;                vector<int> index;                while(left<right){            sum = num[left]+num[right];            if(sum > target)              --right;            else if(sum < target)              ++left;            else{                for(int i = 0; i < length ; ++i){                    if(index.size()==2)                      break;                    if(numbers[i]==num[left])                      index.push_back(i+1);                    else if(numbers[i]==num[right])                      index.push_back(i+1);                                    }                return index;            }                    }                   }};

java 版本-有问题

public class Solution {    public int[] twoSum(int[] numbers, int target) {      int[] num = numbers.clone();      Arrays.sort(num);//unstability sort            int length = numbers.length;      int left = 0;      int right = length -1;      int sum = 0;      int[] index = new int[2];            while(left<right){          sum= num[left] + num[right];                if(sum<target)            ++left;          else if(sum > target)            --right;          else{              int count=0;              for(int i = 0; i < length; i++){                  if(count==2)                    break;                  if(numbers[i]==num[left]){                    index[0]=i+1;                    ++count;                    continue;                  }                  else if(numbers[i]==num[right]){                       index[1]=i+1;                     ++count;                     continue;                  }                                                  }              break;          }      }      return index;    }}

问题在于

  if(numbers[i]==num[left]){

如果两个因子相同，则只改变了left，而没有改变right，所以要用容器

public class Solution {    public int[] twoSum(int[] numbers, int target) {      int[] num = numbers.clone();      Arrays.sort(num);//unstability sort            int length = numbers.length;      int left = 0;      int right = length -1;      int sum = 0;      int[] index = new int[2];      ArrayList<Integer> list = new ArrayList<Integer>();// not int            while(left<right){          sum= num[left] + num[right];                if(sum<target)            ++left;          else if(sum > target)            --right;          else{              list.clear();              for(int i = 0; i < length; i++){                  if(list.size()==2)                    break;                  if(numbers[i]==num[left]){                    list.add(i+1);                  }                  else if(numbers[i]==num[right]){                       list.add(i+1);                  }                                                  }              break;          }      }            int i = 0;      for(Iterator it = list.iterator();it.hasNext();++i){        index[i]=(int)it.next();      }      return index;    }}

借鉴：http://blog.csdn.net/lilong_dream/article/details/19298357

LeetCode Solution(持续更新，java>c++)

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