88. Merge Sorted Array

Given two sorted integer arrays

nums1

and

nums2

, merge

nums2

into

nums1

as one sorted array.

Note:

You may assume that

nums1

has enough space (size that is greater or equal to

m + n)

to hold additional elements from

nums2

. The number of elements initialized in

nums1

and

nums2

are

m

and

n

respectively.

思路：参考Merge Sort，也就是Introduction to Algorithms(2nd Edition) p29的

line 10 - line 17

：

i <- 1
j <- 1
for k <- p to r
do if L[i] <= R[j]
then A[k] <- L[i]
i <- i + 1
else A[k] <- R[j]
j <- j + 1

不过这道题目里面我们是从nums的m + n - 1位向前倒着比较然后merge sort的，这样可以省去额外的空间。最后，考虑如果nums2的长度比nums1长的话，就直接把nums2的后面部分直接逆序贴到nums1上去。

代码如下：

class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int a = m - 1, b = n - 1, i = m + n - 1;

while(a >= 0 && b >= 0){
if(nums1[a] > nums2[b])
nums1[i--] = nums1[a--];
else
nums1[i--] = nums2[b--];
}

while(b >= 0){
nums1[i--] = nums2[b--];
}
}
};

Time: O(m + n).

Space: O(1).

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