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Commit 9ba3f6e

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solve 349.两个数组的交集
1 parent be621a1 commit 9ba3f6e

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3 files changed

+223
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lines changed

‎zh/349.两个数组的交集.1.java‎

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/*
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* @lc app=leetcode.cn id=349 lang=java
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*
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* [349] 两个数组的交集
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*
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* https://leetcode-cn.com/problems/intersection-of-two-arrays/description/
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*
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* algorithms
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* Easy (69.29%)
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* Likes: 200
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* Dislikes: 0
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* Total Accepted: 77.4K
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* Total Submissions: 111K
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* Testcase Example: '[1,2,2,1]\n[2,2]'
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*
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* 给定两个数组,编写一个函数来计算它们的交集。
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*
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*
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*
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* 示例 1:
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*
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* 输入:nums1 = [1,2,2,1], nums2 = [2,2]
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* 输出:[2]
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*
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*
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* 示例 2:
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*
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* 输入:nums1 = [4,9,5], nums2 = [9,4,9,8,4]
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* 输出:[9,4]
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*
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*
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*
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* 说明:
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*
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*
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* 输出结果中的每个元素一定是唯一的。
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* 我们可以不考虑输出结果的顺序。
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*
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*
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*/
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// @lc code=start
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class Solution {
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public int[] intersection(int[] nums1, int[] nums2) {
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Arrays.sort(nums1);
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Arrays.sort(nums2);
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int i = 0, j = 0;
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Set<Integer> set = new HashSet<>();
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while (i < nums1.length && j < nums2.length) {
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if (nums1[i] == nums2[j]) {
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set.add(nums1[i]);
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i++;
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j++;
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} else if (nums1[i] < nums2[j]) {
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i++;
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} else if (nums1[i] > nums2[j]) {
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j++;
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}
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}
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int[] result = new int[set.size()];
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int index = 0;
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for (int num : set) {
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result[index++] = num;
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}
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return result;
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}
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}
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// @lc code=end
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‎zh/349.两个数组的交集.2.java‎

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/*
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* @lc app=leetcode.cn id=349 lang=java
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*
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* [349] 两个数组的交集
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*
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* https://leetcode-cn.com/problems/intersection-of-two-arrays/description/
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*
8+
* algorithms
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* Easy (69.29%)
10+
* Likes: 200
11+
* Dislikes: 0
12+
* Total Accepted: 77.4K
13+
* Total Submissions: 111K
14+
* Testcase Example: '[1,2,2,1]\n[2,2]'
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*
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* 给定两个数组,编写一个函数来计算它们的交集。
17+
*
18+
*
19+
*
20+
* 示例 1:
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*
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* 输入:nums1 = [1,2,2,1], nums2 = [2,2]
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* 输出:[2]
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*
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*
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* 示例 2:
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*
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* 输入:nums1 = [4,9,5], nums2 = [9,4,9,8,4]
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* 输出:[9,4]
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*
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*
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*
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* 说明:
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*
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*
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* 输出结果中的每个元素一定是唯一的。
37+
* 我们可以不考虑输出结果的顺序。
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*
39+
*
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*/
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// @lc code=start
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class Solution {
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public int[] intersection(int[] nums1, int[] nums2) {
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Arrays.sort(nums2);
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Set<Integer> set = new HashSet<>();
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for (int num : nums1) {
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if (!set.contains(num) && bsearch(nums2, num)) {
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set.add(num);
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}
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}
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int[] result = new int[set.size()];
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int index = 0;
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for (int num : set) {
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result[index++] = num;
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}
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return result;
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}
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public boolean bsearch(int[] nums, int target) {
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int left = 0;
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int right = nums.length - 1;
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while (left <= right) {
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int mid = left + (right - left) / 2;
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if (nums[mid] == target) {
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return true;
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} else if (nums[mid] < target) {
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left = mid + 1;
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} else if (nums[mid] > target) {
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right = mid - 1;
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}
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}
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return false;
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}
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}
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// @lc code=end
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‎zh/349.两个数组的交集.java‎

Lines changed: 69 additions & 0 deletions
Original file line numberDiff line numberDiff line change
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/*
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* @lc app=leetcode.cn id=349 lang=java
3+
*
4+
* [349] 两个数组的交集
5+
*
6+
* https://leetcode-cn.com/problems/intersection-of-two-arrays/description/
7+
*
8+
* algorithms
9+
* Easy (69.29%)
10+
* Likes: 200
11+
* Dislikes: 0
12+
* Total Accepted: 77.4K
13+
* Total Submissions: 111K
14+
* Testcase Example: '[1,2,2,1]\n[2,2]'
15+
*
16+
* 给定两个数组,编写一个函数来计算它们的交集。
17+
*
18+
*
19+
*
20+
* 示例 1:
21+
*
22+
* 输入:nums1 = [1,2,2,1], nums2 = [2,2]
23+
* 输出:[2]
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*
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*
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* 示例 2:
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*
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* 输入:nums1 = [4,9,5], nums2 = [9,4,9,8,4]
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* 输出:[9,4]
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*
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*
32+
*
33+
* 说明:
34+
*
35+
*
36+
* 输出结果中的每个元素一定是唯一的。
37+
* 我们可以不考虑输出结果的顺序。
38+
*
39+
*
40+
*/
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// @lc code=start
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class Solution {
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public int[] intersection(int[] nums1, int[] nums2) {
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Set<Integer> set1 = new HashSet<>();
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Set<Integer> set2 = new HashSet<>();
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for (int num : nums1) {
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set1.add(num);
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}
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for (int num : nums2) {
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set2.add(num);
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}
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List<Integer> list = new ArrayList<>();
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for (int num : set2) {
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if (set1.contains(num)) {
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list.add(num);
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}
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}
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int[] result = new int[list.size()];
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for (int i = 0; i < list.size(); i++) {
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result[i] = list.get(i);
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}
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return result;
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}
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}
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// @lc code=end
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