349. Intersection of Two Arrays (Easy) 高频

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

Example:
Givennums1=[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.

Solution 1: HastSet O(m + n); O(min(m, n))

    public int[] intersection(int[] nums1, int[] nums2) {
        Set<Integer> set = new HashSet<>();
        Set<Integer> intersect = new HashSet<>(); //这个hashset也是必须的，对结果去重
        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;
    }

Solution 2: sort + Two Pointers O(m log m + n log n); O(min(m, n))

    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;
    }

Solution 3: sort + Binary Search O(n log n); O(min(m, n))

    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;
    }

Follow Up:

1.给的数组是已经排好序的，求其交集。重复元素可以选择只保留一个。

Two Pointers + HashSet O(m+n); O(min(m, n))

    public int[] intersection(int[] nums1, int[] nums2) {
        Set<Integer> set = new HashSet<>();
        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;
    }

对于一个数组远大于另一个的情况，如何降低复杂度。

假设n远大于m，那遍历nums1，对nums2进行二分搜索。
Binary Search O(mlogn); O(min(m, n))

    public int[] intersection(int[] nums1, int[] nums2) {
        Set<Integer> set = new HashSet<>();
        for (Integer num : nums1) { //O(m)
            if (binarySearch(nums2, num)) { //O(logn)
                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;
    }