算法总结之 在两个排序数组中找到第K小的数

给定两个有序数组arr1 和 arr2 ，再给定一个int K，返回所有的数中第K小的数

要求长度如果分别为 N M，时间复杂度O（log(min{M,N}),额外空间复杂度O（1）

 

解决此题的方法跟之前的求两个数组求中位数的情况，如出一辙~ 非常给力！

 此题目需要分情况讨论： 

    假设长度较短的数组长度 lenS   较长的lenL

    情况1、 K<1  或者 K>lenS+lenL    k值无效

    情况2、 k<=lenS  分别在两数组选择第前 k个数， 然后取其中位数

    情况3、 k>lenL  

package TT;

public class Test13 {

  public static int getUpMedian(int[] a1, int s1, int e1,int[] a2, int s2, int e2){
         int mid1 = 0;
         int mid2 =0;
         int offset = 0;
         while(s1<e1){
             mid1 = (s1+e1) /2;
             mid2 = (s2+e2) /2;
             offset = ((e1-s1+1)&1)^1;
             if (a1[mid1]>a2[mid2]) {
                 e1 = mid1;
                 s2 = mid2+offset;        
            }else if(a1[mid1]<a2[mid2]){
                s1 = mid1 + offset;
                e2= mid2;
            }else {
                return a1[mid1];
            }
             
             
             
         }
         
      return Math.min(a1[s1], a2[s2]);
      
  }    
    
  public static int findKthNum(int[] arr1, int[] arr2, int kth){
      
      
      if(arr1==null || arr2==null){
          throw new RuntimeException("are you ok?");
      }
      if(kth<1 || kth>arr1.length+arr2.length){
          
          throw new RuntimeException("too long");
          
      }
      int[] longs = arr1.length >=arr2.length ? arr1 :arr2;
      int[] shorts = arr1.length <arr2.length ? arr1 :arr2;
      
      int l  = longs.length;
      int s = shorts.length;
      
      if(kth <= s){
          return getUpMedian(shorts, 0, kth-1, longs, 0, kth-1);
      }
      
      if(kth>l){
          if(shorts[kth-l-1]>=longs[l-1]){
              return shorts[kth-l-1];
          }
          if(longs[kth-s-1]>=shorts[s-1]){
              return longs[kth-s-1];
          }
          return getUpMedian(shorts, kth-l, s-1, longs, kth-s, l-1);
          
      }
      
      if (longs[kth-s-1]>=shorts[s-1]) {
           return longs[kth -s -1];
    }
      return getUpMedian(shorts, 0, s-1, longs, kth-s, kth-1);
      
  }
    
  
  public static void main(String[] args ){
      
      int[] a1 = new int[4];
      int[] a2 = new int[4];
       
      a1[0]=0;  a1[1]=1;  a1[2]=2; a1[3]=3;
      a2[0]=4;  a2[1]=5;  a2[2]=5; a2[3]=6;
      
      int kth=4;
      
      int c = findKthNum( a1, a2, kth);
      System.out.println(c);
      
      
      
      
  }
  
  
  
    
}

结果：