LeetCode:Merge Sorted Array
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You may assume that A has enough space (size that is greater or equal to m + n) to hold additional elements from B. The number of elements initialized in A and B are m andn respectively.
这题目很坑,如果A是[]的话,实际上是[0]但是m=0
class Solution: # @param A a list of integers # @param m an integer, length of A # @param B a list of integers # @param n an integer, length of B # @return nothing def merge(self, A, m, B, n): if m==0: for i in range(n):#如果m=0的话,说明A是空,所以把B的值一个一个赋给A,不能A=B不然输出不正确。。。 A[i]=B[i] return elif n!=0:#如果B非空,B是空的话,就不用计算了 curA=m-1 curB=n-1 cur=m+n-1 while (cur>=0):#指向混合后数组最后一个,向前一个一个赋值 if curA>=0:#如果curA变负数了,说明A的已经全部结束,这时要把B剩下的值放到最前面去 if B[curB]>=A[curA]: A[cur]=B[curB] curB-=1 if curB<0:如果curB变负数了,说明B的值全放到A里面去了,则自然结束 return cur-=1 else: A[cur]=A[curA] curA-=1 cur-=1 else: A[cur]=B[curB]#如果curA变负数了,说明A的以及全部结束,这时要把B剩下的值放到最前面去 curB-=1 cur-=1 return
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