leetcode[89] Merge Sorted Array

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