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.
方法1： 开辟新数组进行归并操作。额外的存储空间为O(m+n)；
class Solution { public: void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) { vector<int> res ; int i=0,j=0; while(i<m&&j<n){ if(nums1[i]<nums2[j]){ res.push_back(nums1[i]); i++; } else{ res.push_back(nums2[j]); j++; } } while(i<m){ res.push_back(nums1[i]); i++; } while(j<n){ res.push_back(nums2[j]); j++; } nums1=res; } };方法2： 因为nums1数组有足够的空间，所以利用这点从尾部开始从大到小进行排序。额外的存储空间为O(1)。
class Solution { public: void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) { int i = m - 1; int j = n - 1; int k = m + n - 1; while(i>=0&&j>=0){ if(nums1[i]>nums2[j]){ nums1[k--]=nums1[i--]; } else{ nums1[k--]=nums2[j--]; } } while(j>=0){ nums1[k--]=nums2[j--]; } } };转载于:https://www.cnblogs.com/GoFly/p/5751065.html
