s****j
发帖数: 67 | 1 int a[maxn],b[maxn];
int find(int aleft,int aright,int bleft,int bright,int k) {
if ((aright-aleft+1)+(bright-bleft+1)
cout<<"error"<
return 0;
}
if (aleft>aright)
return b[bleft+k-1];
if (bleft>bright)
return a[aleft+k-1];
int amid=(aleft+aright)>>1;
int bmid=(bleft+bright)>>1;
if (a[amid]<=b[bmid])
if (k<=(amid-aleft)+(bmid-bleft)+1)
return find(aleft,aright,bleft,bmid-1,k);
else
retu... 阅读全帖 |
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t******r
发帖数: 209 | 2 we'd better to do the left sides of A and B through a loop, instead of using
the formula directly, to avoid overflow. The loop for A's left side is
int aLeft = 0;
for (int i = 0; i < arr.length; ++i) {aLeft += i + 1 - arr[i];}
我的理解就是每次循环求一个差直(比较小),这样就不会溢出了 |
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s*******f
发帖数: 1114 | 3 python:
def SameBST(a, b):
if len(a) < 1:
return len(b) < 1;
if len(a) != len(b) or a[0] != b[0]:
return False
aleft = [x for x in a if x < a[0]]
aright = [x for x in a if x > a[0]]
bleft = [x for x in b if x < b[0]]
bright = [x for x in b if x > b[0]]
return SameBST(aleft, bleft) and SameBST(aright, bright)
if __name__ == "__main__":
a = [10, 5, 20, 15, 30]
b = [10, 20, 15, 30, 5]
b1 = SameBST(a, b)
c = [10, 5, 20, 15, 30]
d = ... 阅读全帖 |
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