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  • Intersection of Two Arrays

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

    Example:
    Given nums1 = [1, 2, 2, 1]nums2 = [2, 2], return [2].

    Note:

    • Each element in the result must be unique.
    • The result can be in any order.
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    解法一:

    2 hashset

    public class Solution {
        public int[] intersection(int[] nums1, int[] nums2) {
            Set<Integer> set = new HashSet<>();
            Set<Integer> intersect = new HashSet<>();
            for (int i = 0; i < nums1.length; i++) {
                set.add(nums1[i]);
            }
            for (int i = 0; i < nums2.length; i++) {
                if (set.contains(nums2[i])) {
                    intersect.add(nums2[i]);
                }
            }
            int[] result = new int[intersect.size()];
            int i = 0;
            for (Integer num : intersect) {
                result[i++] = num;
            }
            return result;
        }
    }

    解法二:

    Sort both arrays, use two pointers

    Time complexity: O(nlogn)

    public class Solution {
        public int[] intersection(int[] nums1, int[] nums2) {
            Set<Integer> set = new HashSet<>();
            Arrays.sort(nums1);
            Arrays.sort(nums2);
            int i = 0;
            int j = 0;
            while (i < nums1.length && j < nums2.length) {
                if (nums1[i] < nums2[j]) {
                    i++;
                } else if (nums1[i] > nums2[j]) {
                    j++;
                } else {
                    set.add(nums1[i]);
                    i++;
                    j++;
                }
            }
            int[] result = new int[set.size()];
            int k = 0;
            for (Integer num : set) {
                result[k++] = num;
            }
            return result;
        }
    }

    解法三:

    binary search

    Time complexity: O(nlogn)

    public class Solution {
        public int[] intersection(int[] nums1, int[] nums2) {
            Set<Integer> set = new HashSet<>();
            Arrays.sort(nums2);
            for (Integer num : nums1) {
                if (binarySearch(nums2, num)) {
                    set.add(num);
                }
            }
            int i = 0;
            int[] result = new int[set.size()];
            for (Integer num : set) {
                result[i++] = num;
            }
            return result;
        }
        
        public boolean binarySearch(int[] nums, int target) {
            int low = 0;
            int high = nums.length - 1;
            while (low <= high) {
                int mid = low + (high - low) / 2;
                if (nums[mid] == target) {
                    return true;
                }
                if (nums[mid] > target) {
                    high = mid - 1;
                } else {
                    low = mid + 1;
                }
            }
            return false;
        }
    }

    reference:https://discuss.leetcode.com/topic/45685/three-java-solutions

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  • 原文地址:https://www.cnblogs.com/hygeia/p/5697748.html
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