两个算法--Day5



/*
 * 1.Given an array of integers, every element appears twice except for one. Find that single one.
 * (Note:Your algorithm should have a linear runtime complexity.
 * Could you implement it without using extra memory?)
 * 2.Given a string, find the first non-repeating character in it
 *   and return it's index. If it doesn't exist, return -1.
 *   (Note: You may assume the string contain only lowercase letters.)
 * 3.Given an array of integers, return indices of the two numbers such that they add up to a specific target.
 *   You may assume that each input would have exactly one solution, and you may not use the same element twice.
 */
package algorithm;

public class Day5 {

 public static void main(String[] args) {
  // TODO Auto-generated method stub
  Solution sl=new Solution();
  int arr[]={1,3,4,4,3,5,1};
  sl.singleNum(arr);
  sl.firstUni("leecode lee code 0 ,");
  int arr1[]={2,4,6,7,1,0};
  sl.twoSum(arr1, 10);

 }

}

class Solution
{
 
 public void singleNum(int arr[])
 {
  int result=arr[0];
  for(int i=1;i<arr.length;i++)
  {
   result ^=arr[i];
  }
  System.out.println(result);
 }
 public void firstUni(String s)
 {
  int arr[]=new int[255];
  for(int i=0;i<s.length();i++)
  {
   arr[s.charAt(i)]++;
  }
  for(int j=0;j<s.length();j++)
  {
   if(arr[s.charAt(j)]==1)
   {
    System.out.println(j);
    break;
   }
   if(j==s.length()-1)
   {
    System.out.println(-1);
   }
  }
 }
 public void twoSum(int arr[],int target)
 {
  
  for (int i=0;i<arr.length;i++)
  {
   for(int j=i+1;j<arr.length;j++)
   {
    if(arr[i]+arr[j]==target)
    {
     System.out.println(i+" "+j);
     break;
    }
   }
  }
 }
}


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