leetcode 第一题 Two Num java

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最新推荐文章于 2025-04-21 22:09:55 发布

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分类专栏： leetcode 文章标签： leetcode 快速排序

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本文介绍了一种解决两数之和问题的方法，通过使用哈希表快速查找数组中能够与当前元素相加等于目标值的另一个元素。此外，还提供了一个基于快排的解决方案，虽然效率较低但展示了另一种解决问题的思路。

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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、将数组放入hash表，遍历数组，在hash表中查找另一个数。hash查找速度较快

      2、利用快排先将数组排序，双向遍历，找到两数之和为target的两个数m,n ，再遍历原数组，找到m,n的下标

----------转载---------

<pre name="code" class="java">public class Solution {
    public int[] twoSum(int[] numbers, int target) {
                int []result = new int[2];
        java.util.HashMap<Integer, Integer> table = new java.util.HashMap<Integer, Integer>(200000);  
        for (int i = 0; i < numbers.length; i++) {
            table.put(numbers[i], i+1);
        }
        for (int i = 0; i < numbers.length; i++) {
            if(table.get(target-numbers[i])!=null&&table.get(target-numbers[i])!=i+1){
                result[0]=i+1;
                result[1]=table.get(target-numbers[i]);
                break;
            }
        }
        return result;
    }
}

</pre><pre name="code" class="java">---------my coding    time limit exceeded ---------

public class Solution1 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int[] num={2,3,4,5,6,8,9};
		Solution1 s=new Solution1();
		int[] indexs= s.twoSum(num, 10);
	       
		System.out.println(indexs[0]+" "+indexs[1]);
	}
	
	
	    public int[] twoSum(int[] numbers, int target) {
	        int length=numbers.length;
	        int[] sortNum=numbers;
	        sortNum=quickSort(sortNum,0,length-1);
	        
	        int[] index=new int[2];
	        int split1=0,split2=0,index1=0,index2=0;
	        for(int i=0;i<length-1;i++){
	            if(2*sortNum[i]<=target && 2*sortNum[i+1]>target){
	                split1=i;    
	            }
	            if(sortNum[i]>target){
	            	split2=i;
	            	break;         	
	            }
	        }
	        if(split2==0){
	        	split2=length;
	        }
	        
	        index1=0;index2=split1;
	        while(index1<split1 && index2<split2 ){
	        	
	            if(sortNum[index1]+sortNum[index2] <target){
	            	index1++;
	                if(index1==split1){
	                	index1=0;
	                	index2++;  	
	                }
	                
	            }else if(sortNum[index1]+sortNum[index2] >target){
	            	index1=0;
	            	index2++;
	            }else{
	                index[0]=sortNum[index1];
	                index[1]=sortNum[index2];
	                break;
	            }
	        }
	        
	        if(index[0]!=0 && index[1]!=0){
		        for(int i=0;i<numbers.length;i++){
			           if(numbers[i]==index[0])
			        	   index[0]=i+1;
			           if(numbers[i]==index[1])
			            	index[1]=i+1;
			    }
		        if(index[0]>index[1]){
		            int temp=index[0];
		            index[0]=index[1];
		            index[1]=temp;
		        }
	        }
	       return index;
	        
	    }
	    
	    int[] quickSort(int[] num,int left,int right){
	        
	        if(left<right){
	            int i=left,j=right,x=num[left];
	            while(i<j){
	                while(i<j && num[j]>=x)
	                    j--;
	                if(i<j)
	                    num[i++]=num[j];
	                    
	                while(i<j && num[i]<x)
	                    i++;
	                if(i<j)
	                    num[j--]=num[i];
	            }
	            num[i]=x;
	            
	            quickSort(num,left,i-1);
	            quickSort(num,i+1,right);
	        }
	        
	        return num;
	    }
	 

	

}

</pre><pre name="code" class="java">