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  • 349. Intersection of Two Arrays

    Given two arrays, write a function to compute their intersection.

    Example 1:

    Input: nums1 = [1,2,2,1], nums2 = [2,2]
    Output: [2]
    

    Example 2:

    Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
    Output: [9,4]

    Note:

    • Each element in the result must be unique.
    • The result can be in any order.
    //Approach 1: HashSet
    //Time: O(n), Space: O(n) 
       public int[] intersection(int[] nums1, int[] nums2) {
            if (nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0) {
                return new int[0];
            }
            
            HashSet<Integer> result = new HashSet<Integer>();
            HashSet<Integer> set = new HashSet<Integer>();
            
            for (int num : nums1) {
                set.add(num);
            }
            
            for (int num : nums2) {
                if (set.contains(num)) {
                    result.add(num);
                }
            }
            
            int[] ans = new int[result.size()];
            int i = 0;
            
            for(int ele : result) {//注意:遍历set没法用index assign给array,只能i递增
                ans[i] = ele;
                i++;
            }
            
            return ans;
        }
    
    //Approach 2: Sort + Two pointer
    //Time: O(nlogn), Space: O(1)
        public int[] intersection(int[] nums1, int[] nums2) {
            if (nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0) {
                return new int[0];
            }
            
            HashSet<Integer> result = new HashSet<Integer>();
            Arrays.sort(nums1);
            Arrays.sort(nums2);
            int i = 0;
            int j = 0;
            
            while (i < nums1.length && j < nums2.length) {
                if (nums1[i] == nums2[j]) {
                    result.add(nums1[i]);
                    i++;
                    j++;
                } else if (nums1[i] < nums2[j]) {
                    i++;
                } else {
                    j++;
                }
            }
            
            int[] ans = new int[result.size()];
            int k = 0;
            
            for(int ele : result) {
                System.out.println(ele);
                ans[k] = ele;
                k++;
            }
            
            return ans;
        }
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  • 原文地址:https://www.cnblogs.com/jessie2009/p/9771946.html
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