@ysner
2021-10-26T13:54:20.000000Z
字数 1842
阅读 890
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第一次提交的代码:(归并排序)
class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int[] num=new int[3000];
int head1,head2,head,m=nums1.length,n=nums2.length;//求数组长度:.length就可,不要括号!!!
head1=head2=head=0;
while(head1<m&&head2<n)
{
if(nums1[head1]>nums2[head2])
{
num[head]=nums2[head2];
head++;
head2++;
}
else
{
num[head]=nums1[head1];
head++;
head1++;
}
}
while(head1<m)
{
num[head]=nums1[head1];
head++;
head1++;
}
while(head2<n)
{
num[head]=nums2[head2];
head++;
head2++;
}
int t=(n+m)/2;
return ((n+m)%2==1)?num[t]:((num[t-1]+num[t])/2.0); //注意想的是第n个数,写的编号是n-1
}
}
第二次提交的代码:(归并排序+常数优化)
class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int head1,head2,head,m=nums1.length,n=nums2.length,t,Now=0,Las=0;
head1=head2=head=0;
t=(n+m)/2;
boolean tag=((n+m)%2==1);
while(head1<m&&head2<n)
{
if(nums1[head1]>nums2[head2])
{
Las=Now;Now=nums2[head2];
if(head==t) return tag?Now:((Now+Las)/2.0);
head++;head2++;
}
else
{
Las=Now;Now=nums1[head1];
if(head==t) return tag?Now:((Now+Las)/2.0);
head++;head1++;
}
}
while(head1<m)
{
Las=Now;Now=nums1[head1];
if(head==t) return tag?Now:((Now+Las)/2.0);
head++;head1++;
}
while(head2<n)
{
Las=Now;Now=nums2[head2];
if(head==t) return tag?Now:((Now+Las)/2.0);
head++;head2++;
}
return -1;
}
}
第三次提交的代码:(二分查找)
class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
if(nums1.length>nums2.length)
{
int[] tmp=nums1;
nums1=nums2;
nums2=tmp;//用小数组二分,减小二分范围
}
int m=nums1.length,n=nums2.length;
int tot=(n+m+1)/2;
int L=0,R=m;//二分nums1出了多少个数,结果为L
while(L<R)
{
int mid=(L+R+1)/2,j=tot-mid;//二分注意:不加1的话会在R-L=1时出死循环
if(nums1[mid-1]<=nums2[j]) L=mid;
else R=mid-1;
}
int i=L,j=tot-i;
int s1L=(i==0)?Integer.MIN_VALUE:nums1[i-1],
s1R=(i==m)?Integer.MAX_VALUE:nums1[i],
s2L=(j==0)?Integer.MIN_VALUE:nums2[j-1],
s2R=(j==n)?Integer.MAX_VALUE:nums2[j];//特殊处理分割线恰好在数组边界的情况
if((n+m)%2==1) return Math.max(s1L,s2L);
else return (Math.max(s1L,s2L)+Math.min(s1R,s2R))/2.0;
}
}