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]
模拟法 O(n)
对nums1和nums2从m, n开始往前扫,挑大的数放到nums1最后面的位置,三根指针都从后往前。
从后往前能避免nums1和nums2的还没处理过的候选数据不会被覆盖掉。从前往后会有这个隐患。
细节:
1.最后如果两个数组中还有剩下的直接接着粘这一节。
实现:
class Solution { public void merge(int[] nums1, int m, int[] nums2, int n) { int i = m - 1, 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 (i >= 0) { nums1[k--] = nums1[i--]; } while (j >= 0) { nums1[k--] = nums2[j--]; } } }