https://leetcode.com/problems/median-of-two-sorted-arrays/#/description
除了要想到通过寻找第k大的数来优化算法外，此题还有非常变态的边界处理
class Solution {
public:
int fuck(int sx,int x,vector<int>& a,int sy,int y,vector<int>& b,int k)
{
if(x-sx>y-sy)
return fuck(sy,y,b,sx,x,a,k);
if(x==sx)
return b[sy+k-1];
if(k==1)
{
if(a[sx]<b[sy])
return a[sx];
else
return b[sy];
}
int apos=min(sx+k/2-1,x-1);
int bpos=sy+k-apos+sx-2;
//printf("%d %d %d %d %d
",apos,bpos,a[apos],b[bpos],k);
if(a[apos]==b[bpos])
return a[apos];
if(a[apos]<b[bpos])
return fuck(apos+1,x,a,sy,y,b,k-(apos-sx+1));
else
return fuck(sx,x,a,bpos+1,y,b,k-(bpos-sy+1));
}
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
int len1=nums1.size();
int len2=nums2.size();
if((len1+len2)&1)
return fuck(0,len1,nums1,0,len2,nums2,(len1+len2)/2+1);
else
{
return (fuck(0,len1,nums1,0,len2,nums2,(len1+len2)/2)+
fuck(0,len1,nums1,0,len2,nums2,(len1+len2)/2+1))/2.0;
}
}
};
嘛，找第k大的做法是这样地
Assume that the number of elements in A and B are both larger than k/2, and if we compare the k/2-th smallest element in A(i.e. A[k/2-1]) and the k-th smallest element in B(i.e. B[k/2 - 1]), there are three results:
(Becasue k can be odd or even number, so we assume k is even number here for simplicy. The following is also true when k is an odd number.)
A[k/2-1] = B[k/2-1]
A[k/2-1] > B[k/2-1]
A[k/2-1] < B[k/2-1]
if A[k/2-1] < B[k/2-1], that means all the elements from A[0] to A[k/2-1](i.e. the k/2 smallest elements in A) are in the range of k smallest elements in the union of A and B. Or, in the other word, A[k/2 - 1] can never be larger than the k-th smalleset element in the union of A and B.
Why?
We can use a proof by contradiction. Since A[k/2 - 1] is larger than the k-th smallest element in the union of A and B, then we assume it is the (k+1)-th smallest one. Since it is smaller than B[k/2 - 1], then B[k/2 - 1] should be at least the (k+2)-th smallest one. So there are at most (k/2-1) elements smaller than A[k/2-1] in A, and at most (k/2 - 1) elements smaller than A[k/2-1] in B.So the total number is k/2+k/2-2, which, no matter when k is odd or even, is surly smaller than k(since A[k/2-1] is the (k+1)-th smallest element). So A[k/2-1] can never larger than the k-th smallest element in the union of A and B if A[k/2-1]<B[k/2-1];
Since there is such an important conclusion, we can safely drop the first k/2 element in A, which are definitaly smaller than k-th element in the union of A and B. This is also true for the A[k/2-1] > B[k/2-1] condition, which we should drop the elements in B.
When A[k/2-1] = B[k/2-1], then we have found the k-th smallest element, that is the equal element, we can call it m. There are each (k/2-1) numbers smaller than m in A and B, so m must be the k-th smallest number. So we can call a function recursively, when A[k/2-1] < B[k/2-1], we drop the elements in A, else we drop the elements in B.
We should also consider the edge case, that is, when should we stop?
1. When A or B is empty, we return B[k-1]( or A[k-1]), respectively;
2. When k is 1(when A and B are both not empty), we return the smaller one of A[0] and B[0]
3. When A[k/2-1] = B[k/2-1], we should return one of them
In the code, we check if m is larger than n to garentee that the we always know the smaller array, for coding simplicy.
之后调试了很久的边界。。总结额外要主意几种情况
1.当k/2大于当前A数组长度时
2.当k/2大于当前A数组长度时
3.当k/2大于当前A数组长度时
嘛，写了几遍wa都怎么想也想不到，试很久才试出来的。其实如果要独自构思整个思路，还要考虑到A is empty和k==1这两种情况，都挺难想的