[Leetcode][第题][JAVA][两个数组的交集 II1][双指针][HashMap]

【问题描述】[中等]

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【解答思路】

1. 哈希映射

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复杂度分析
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class Solution {public int[] intersect(int[] nums1, int[] nums2) {if (nums1.length > nums2.length) {return intersect(nums2, nums1);}Map<Integer, Integer> map = new HashMap<Integer, Integer>();for (int num : nums1) {int count = map.getOrDefault(num, 0) + 1;map.put(num, count);}int[] intersection = new int[nums1.length];int index = 0;for (int num : nums2) {int count = map.getOrDefault(num, 0);if (count > 0) {intersection[index++] = num;count--;if (count > 0) {map.put(num, count);} else {map.remove(num);}}}return Arrays.copyOfRange(intersection, 0, index);}
}

 public int[] intersect(int[] nums1, int[] nums2) {HashMap<Integer, Integer> map = new HashMap<>();List<Integer> list = new ArrayList<>();for(int num : nums1) {if(!map.containsKey(num)) map.put(num, 1);else map.put(num, map.get(num) + 1);}for(int num : nums2) {if(map.containsKey(num)) {map.put(num, map.get(num) - 1);if(map.get(num) == 0) map.remove(num);list.add(num);}}int[] res = new int[list.size()];for(int i = 0; i < list.size(); i++) {res[i] = list.get(i);}return res;}

2. 双指针

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复杂度分析

class Solution {public int[] intersect(int[] nums1, int[] nums2) {Arrays.sort(nums1);Arrays.sort(nums2);int length1 = nums1.length, length2 = nums2.length;int[] intersection = new int[Math.min(length1, length2)];int index1 = 0, index2 = 0, index = 0;while (index1 < length1 && index2 < length2) {if (nums1[index1] < nums2[index2]) {index1++;} else if (nums1[index1] > nums2[index2]) {index2++;} else {intersection[index] = nums1[index1];index1++;index2++;index++;}}return Arrays.copyOfRange(intersection, 0, index);}
}

  public int[] intersect(int[] nums1, int[] nums2) {Arrays.sort(nums1);Arrays.sort(nums2);List<Integer> list = new ArrayList<>();int p1 = 0, p2 = 0;while(p1 < nums1.length && p2 < nums2.length) {if(nums1[p1] < nums2[p2]) p1++;else if(nums1[p1] > nums2[p2]) p2++;else {list.add(nums1[p1]);p1++;p2++;}}int[] res = new int[list.size()];for(int i = 0; i < res.length; i++) res[i] = list.get(i);return res;}

3. 辅助数组(不推荐)

时间复杂度：O(N^2) 空间复杂度：O(N)

    public int[] intersect(int[] nums1, int[] nums2) {int len1 = nums1.length;int len2 = nums2.length;List<Integer> ans = new ArrayList<>();if(len1<=len2){ans = cross(len1,len2,nums1,nums2);}else{//len1>len2ans = cross(len2,len1,nums2,nums1);}int[] res = new int[ans.size()];for(int i = 0; i < res.length; i++) res[i] = ans.get(i);return res;}public List<Integer> cross(int shao,int duo,int[] nums1,int[] nums2 ){List<Integer> ans = new ArrayList<>();boolean[] a = new boolean[duo]; for(int num:nums1){for(int i = 0;i<duo;i++){if(num == nums2[i] && !a[i]){ans.add(num);a[i]=true;break;}}}return ans;}

【总结】

1.函数返回一个长度不确定的数组（int[ ]）

方法一：List转Int数组（逐个复制）

List<Integer> list = new ArrayList<Integer>();
//LinkedList<Integer> list = new LinkedList<Integer>();list.add(1);list.add(2);int count = list.size();int[] aux = new int[count];for(int i = 0; i < count; i++){aux[i] = list.poll();}return aux;

方法二：使用函数Arrays.copyOfRange

int length1 = nums1.length, length2 = nums2.length;
int[] intersection = new int[Math.min(length1, length2)];
return Arrays.copyOfRange(intersection, 0, index);

2.交集问题双指针 HashMap

3.String[] 与 List < String > 互转

		String[] strings1 = {"a", "b", "c"};// String[] 转 List<String>List<String> list3 = Arrays.asList(strings1);// List<String> 转 String[]String[] strings2 = list3.toArray(new String[0]);