Array——LeetCode——Merge Sorted Array

【学到的知识点——当有两个判断时，谁在外面谁在里面】
1、两个判断
判断一：a和b是否会超出m和n
判断二：nums1[a]大，还是nums2[b]大
2、很显然判断一的优先级大于判断二
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【反思】
1、审题，题目说了nums1有足够的空间
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【别人的Java解法代码】

 1     public void merge(int[] nums1, int m, int[] nums2, int n) {
 2         int[] nums = new int[m+n];
 3         int a = 0;
 4         int b = 0;
 5         for (int i = 0; i < m + n; i++) {
 6             if (a < m && b < n) {
 7                 if (nums1[a] > nums2[b]) {
 8                     nums[i] = nums2[b];
 9                     b++;
10                 } else {
11                     nums[i] = nums1[a];
12                     a++;
13                 }
14             } else if ( a < m && b == n) {
15                 nums[i] = nums1[a];
16                 a++;
17             } else if ( a == m && b < n) {
18                 nums[i] = nums2[b];
19                 b++;
20             } else
21                 return;
22         }
23         for (int i = 0; i < m + n; i++) {
24             nums1[i] = nums[i];
25         }
26     }

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【自己的Java解法代码】

原文地址：https://www.cnblogs.com/Dbbf/p/9639389.html

时间： 2024-08-09 06:20:51

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