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leetcode[89] Merge Sorted Array

2014-11-22 01:07 344 查看
合并两个有序数组,放在A中,A中的空间足够。

Given two sorted integer arrays A and B, merge B into A as one sorted array.

Note:
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.

从后往前放,比较A和B的从后往前的,比较大的放在m+n-1的位置。然后一直减减。

class Solution {
public:
void merge(int A[], int m, int B[], int n)
{while(m > 0 && n > 0)
{
if (A[m-1] > B[n-1])
{
A[m+n-1] = A[m-1];
m--;
}
else
{
A[m+n-1] = B[n-1];
n--;
}
}
if (m == 0)
while(n > 0)
{
A[n-1] = B[n-1];
n--;
}
return ;
}
};


也可以写短一点:

class Solution {
public:
void merge(int A[], int m, int B[], int n)
{
int nowA = m - 1, nowB = n - 1, last = n + m - 1;
while(nowA >= 0 && nowB >=0)
{
A[last--] = A[nowA] > B[nowB] ? A[nowA--] : B[nowB--];
}
while(nowB >= 0)
A[last--] = B[nowB--];
return ;
}
};


2015/03/31:

class Solution {
public:
void merge(int A[], int m, int B[], int n) {
int i = m - 1, j = n - 1, k = m + n - 1;
while(i >=0 && j >=0){
A[k--] = A[i] > B[j] ? A[i--] : B[j--];
}
while(i >= 0){
A[k--] = A[i--];  // 多余了
}
while(j >= 0){
A[k--] = B[j--];
}
}
};


原来多余的写了一部分 A的不用赋值给A可以,因为就是A啊

python:

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(void)
def merge(self, A, m, B, n):
i, j, k= m -1, n -1, m + n -1
while i >=0 and j >=0:
if A[i] > B[j]:
A[k] = A[i]
i -= 1
else:
A[k] = B[j]
j -= 1
k -= 1
while j >= 0:
A[k] = B[j]
k -= 1
j -= 1
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