leetcode350~Intersection of Two Arrays II
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Given two arrays, write a function to compute their intersection.
Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].
Note:
Each element in the result should appear as many times as it shows in both arrays.
The result can be in any order.
方法一,两个指针,直接遍历两个数组,逐个比较,有重复的则放到新的数组里面,移动两个指针,否则移动其中一个数组值较小的指针,继续比较。
方法二,采用map集合记录数组值出现的次数,然后再与另外一个数组比较。
public class IntersectionofTwoArraysII { //遍历两个数组 逐个比较 public int[] intersect2(int[] nums1, int[] nums2) { int i=0,j=0,k=0; Arrays.sort(nums1); Arrays.sort(nums2); ArrayList<Integer> res = new ArrayList<>(); while(i<nums1.length && j<nums2.length) { if(nums1[i]==nums2[j]) { res.add(nums1[i]); i++; j++; k++; } else if(nums1[i]<nums2[j]) { i++; } else { j++; } } int[] result = new int[res.size()]; for(int n=0;i<res.size();i++) { result[n] = res.get(n); } return result; } //使用map集合记录其中一个数组值出现的次数,然后再与另外一个数组进行比较 public int[] intersect(int[] nums1,int[] nums2) { Map<Integer,Integer> map = new HashMap<>(); for(int i=0;i<nums1.length;i++) { if(map.containsKey(nums1[i])) { map.put(nums1[i], map.get(nums1[i])+1); } else { map.put(nums1[i], 1); } } List<Integer> list = new ArrayList<>(); for(int i=0;i<nums2.length;i++) { if(map.containsKey(nums2[i]) && map.get(nums2[i])>0) { list.add(nums2[i]); map.put(nums2[i], map.get(nums2[i])-1); } } int result[] = new int[list.size()]; for(int i = 0; i < list.size(); ++i) result[i] = list.get(i); return result; }}
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