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.
題目標籤：Array
題目給了我們兩個有序陣列，讓我們merge sort，而且nums1 有足夠的空間讓我們存放 兩個array。既然是要放到nums1裡，那麼我們如果比較nums1 和nums2 的數字，如果是nums2 的數字小的話，就要把nums2 的數字存放進nums1，我們還要另外存放nums1的數字，這樣就需要額外space了。所以，我們可以從nums1的end 開始遍歷回start， 然後設兩個pointers 分別指向 兩個array的最後一個數字，這樣可以比完大小後直接存放進nums1。注意最後還需要檢查一下，nums2是否還有剩下的數字，有的話都需要存進nums1。
Java Solution:
Runtime beats 38.77%
完成日期：04/05/2017
關鍵詞：Array
關鍵點：Merge sort；Two Pointers；從end遍歷回start
public class Solution
{
public void merge(int[] nums1, int m, int[] nums2, int n)
{
/* if m = 0, meaning nums1's elements are all done. Need one more while loop after this
to take care of left elements of nums2.
if n = 0, meaning nums2's elements are done, the rest of nums1's elements are in the
right place. No need to take care of them.
*/
while(m > 0 && n >0)
{
if(nums1[m-1] > nums2[n-1]) // if number 1 > number 2
{
// save nums1's element into nums1's last "empty" spot.
nums1[m+n-1] = nums1[m-1];
m--;
}
else // if number 1 <= number 2
{
// save nums2's element into nums1's last "empty" spot
nums1[m+n-1] = nums2[n-1];
n--;
}
} // check if nums2's elements are not finished.
while(n > 0)
{
// save nums2's rest elements into nums1
nums1[m+n-1] = nums2[n-1];
n--;
}
}
}
參考資料：
http://www.cnblogs.com/springfor/p/3872640.html
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