问题描述:
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 greater or 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]
解题思路:
我们可以用一个临时数组来存储排序后的数组,最后将这个数组的值赋给nums1。
还可以从后往前填充,因为后面是空的,不会影响到前面的数字.
代码:
class Solution { public: void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) { vector<int> temp(m+n, 0); int i = 0, j = 0,idx = 0; while(i < m && j < n){ if(nums1[i] <= nums2[j]){ temp[idx++] = nums1[i++]; }else{ temp[idx++] = nums2[j++]; } } while(i < m) temp[idx++] = nums1[i++]; while(j < n) temp[idx++] = nums2[j++]; i = 0; while(i < m+n){ nums1[i] = temp[i++]; } } };
从后向前填充:
class Solution { public: void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) { int i = m-1, j = n-1, idx = m+n-1; while(j >= 0){ nums1[idx--] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--]; } } };