二分。情况讨论
因为数组有序,所以能够考虑用二分。通过二分剔除掉肯定不是第k位数的区间。如果数组A和B当前处理的下标各自是mid1和mid2。则
1、假设A[mid1]<B[mid2],
①、若mid1+mid2+2==k(+2是由于下标是从0開始的),则
mid1在大有序数组中下标肯定小于k,所以能够排除[0,mid1]。此外。B[mid2]下标大于或等于k。能够排除[mid2+1,n];
②、若mid1+mid2+2<k,则
mid1在大有序数组中下标肯定小于k,所以能够排除[0,mid1]
③、若mid1+mid2+2>k,则
B[mid2]下标大于k,能够排除[mid2,n];
2、假设A[mid1]<B[mid2]情况相符,仅仅是下标改变。
这些操作处理完后。可能一个数组被排除了,即满足lowX>highX。此时仅仅需对还有一个数组进行二分,同一时候二分其元素在还有一个数组中的下标,确定全局下标,终于通过推断全局下标与k的关系。确定是否为第k数
class Solution {
public:
int findPos(int* p,int n,int x){
int low=0,high=n-1,mid;
while(low<=high){
mid=(low+high)>>1;
if(p[mid]<=x)low=mid+1;
else high=mid-1;
}
return low;
}
double findK(int a[], int m, int b[], int n,int k){
int mid1,mid2,low1=0,low2=0,high1=m-1,high2=n-1,x;
while(low1<=high1&&low2<=high2){
mid1=(high1+low1)>>1;
mid2=(high2+low2)>>1;
if(a[mid1]<b[mid2]){
if(mid1+mid2+2==k){
low1=mid1+1;
high2=mid2;
}
else if(mid1+mid2+2<k){
low1=mid1+1;
}
else high2=mid2-1;
}
else{
if(mid1+mid2+2==k){
low2=mid2+1;
high1=mid1;
}
else if(mid1+mid2+2<k){
low2=mid2+1;
}
else high1=mid1-1;
}
}
if(low1<=high1){
// if(low1==high1)return a[low1];
while(low1<=high1){
mid1=(low1+high1)>>1;
x=findPos(b,n,a[mid1]);
if(x+mid1+1==k)return a[mid1];
else if(x+mid1<k)low1=mid1+1;
else high1=mid1-1;
}
return low1>=m?a[m-1]:a[low1];
}
else {
// if(low2==high2)return b[low2];
while(low2<=high2){
mid2=(low2+high2)>>1;
x=findPos(a,m,b[mid2]);
if(x+mid2+1==k)return b[mid2];
else if(x+mid2<k)low2=mid2+1;
else high2=mid2-1;
}
return low2>=n? a[n-1]:b[low2];
}
}
double findMedianSortedArrays(int a[], int m, int b[], int n) {
int k=m+n;
if(k&1){
return findK(a,m,b,n,k/2+1);
}
else{
return (findK(a,m,b,n,k/2)+findK(a,m,b,n,k/2+1))/2.0;
}
}
};