[LeetCode] #350 两个数组的交集 II

给定两个数组，编写一个函数来计算它们的交集。

输入：nums1 = [1,2,2,1], nums2 = [2,2]

输出：[2,2]

与[LeetCode] #349 两个数组的交集相比，不需要去重

因此将排序法中去重判断去掉

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[length1 + length2];
        int index = 0, index1 = 0, index2 = 0;
        while (index1 < length1 && index2 < length2) {
            int num1 = nums1[index1], num2 = nums2[index2];
            if (num1 == num2) {
                intersection[index++] = num1;
                index1++;
                index2++;
            } else if (num1 < num2) {
                index1++;
            } else {
                index2++;
            }
        }
        return Arrays.copyOfRange(intersection, 0, index);
    }
}

或者使用HashMap

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

知识点：无

总结：无