lc88:Merge Sorted Array

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:

The number of elements initialized in nums1 and nums2 are m and n respectively. You may assume that nums1 has enough space (size that is equal to m + n) to hold additional elements from nums2. Example:

Input: nums1 = [1,2,3,0,0,0], m = 3 nums2 = [2,5,6], n = 3

Output: [1,2,2,3,5,6]

Constraints:

-10^9 <= nums1[i], nums2[i] <= 10^9 nums1.length == m + n nums2.length == n

解题思路： 因为nums1长度为nums1和nums2 >0的值的总和，所以双指针，倒叙比较插入

代码：python3

class Solution:
    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
        """
        Do not return anything, modify nums1 in-place instead.
        """
        p=m-1
        q=n-1
        r=m+n-1
        while p>=0 and q>=0:
            if nums1[p]>nums2[q]:
                nums1[r]=nums1[p]
                p=p-1
                r=r-1
            else:
                nums1[r]=nums2[q]
                q=q-1
                r=r-1   
        nums1[:q+1]=nums2[:q+1]
if __name__ == '__main__':
    print(Solution().merge([1,2,3,0,0,0],3,[2,5,6],3))    

空间复杂度：O(1),时间复杂度O(m+n)

java 题解 因为nums1，nums2都是排好序的 nums1的最后一位开始逆序添加

class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
		 int i1=m-1;
		 int i2=n-1;
                int cur = m+n-1;
                //i2<0时，说明整个nums1数组排好序了
                 while(i2>=0){
                          //i1<0,可能还需要把nums2剩余元素添加进来
                          while(i1>=0 && nums1[i1]>nums2[i2]){
                                nums1[cur--]=nums1[i1--];
                          }else{
                               nums1[cur--]=nums2[i2--];
                          }
                }
		
	   }
}

空间复杂度：O(1),时间复杂度O(m+n)