LeetCode 88 — Merge Sorted Array（C++ Java Python）

题目：http://oj.leetcode.com/problems/merge-sorted-array/

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 and nrespectively.
题目翻译：

给定两个有序整数数组A和B，把B并入A成为一个有序数组。

注意：

可以假设A有足够的空间（大小大于等于m + n）来保存来自B的额外元素。A和B的初始元素个数分别为m和n。

分析：

        从后向前考虑。如果A中的先遍历完，应把B中剩下的元素复制到A中。
C++实现：

class Solution {
public:
void merge(int A[], int m, int B[], int n) {
if(n == 0)
{
return;
}

int idx = m + n - 1;
int i = m - 1;
int j = n - 1;

while(i >= 0 && j >= 0)
{
if(A[i] >= B[j])
{
A[idx--] = A[i--];
}
else
{
A[idx--] = B[j--];
}
}

if(i == -1)
{
while(j >= 0)
{
A[j] = B[j];
j--;
}
}
}
};

Java实现：

public class Solution {
public void merge(int A[], int m, int B[], int n) {
int idx = m + n - 1;
int i = m - 1;
int j = n - 1;

while (i >= 0 && j >= 0) {
if (A[i] >= B[j]) {
A[idx--] = A[i--];
} else {
A[idx--] = B[j--];
}
}

if (i == -1) {
while (j >= 0) {
A[j] = B[j];
j--;
}
}
}
}

Python实现：

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):
idx = m + n - 1
i = m - 1
j = n - 1

while i >= 0 and j >= 0:
if A[i] >= B[j]:
A[idx] = A[i]
i -= 1
else:
A[idx] = B[j]
j -= 1

idx -= 1

if i == -1:
while j >= 0:
A[j] = B[j]
j -= 1

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