Merge Sorted Array--LeetCode
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1.题目
Merge Sorted Array
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.
2.题意
合并升序数组到nums1中
3.分析
借助3个index,分别对应数组nums1,数组nums2,和结果数组
遍历nums1和nums2,将其中较大值放入结果数组即可
由于结果放在nums1中,所以只能从后往前扫描,
如果从前往后扫描可能会覆盖掉未检查的元素
时间复杂度为O(m+n),m和n分别是两个数组的长度,空间复杂度是O(1)
注意数组尾元素为nums1[m-1]而非nums1[m],不要从m开始遍历
循环的判断条件是idx1 >= 0 && idx2 >= 0而非idx1 > 0 && idx2 > 0
对应的数组名不要混淆,不要遗忘两边index--
的操作
4.代码
class Solution {public: void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) { if(m < 0 || n < 0) return; int idx1 = m - 1; int idx2 = n - 1; int index = m + n - 1; while(idx1 >= 0 && idx2 >= 0) { if(nums1[idx1] > nums2[idx2]) nums1[index--] = nums1[idx1--]; else nums1[index--] = nums2[idx2--]; } while(idx2 >= 0) nums1[index--] = nums2[idx2--]; return; }};
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