350. 两个数组的交集 II

2020-02-02  本文已影响0人  Andysys
    public int[] intersect(int[] nums1, int[] nums2) {
        if (nums1.length == 0 || nums2.length == 0) {
            return new int[0];
        }
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        List<Integer> list = new ArrayList<>();
        int p1 = 0;
        int p2 = 0;
        while(p1 < nums1.length && p2 < nums2.length) {
            if (nums1[p1] < nums2[p2]) {
                p1++;
            } else if (nums1[p1] > nums2[p2]) {
                p2++;
            } else {
                list.add(nums1[p1]);
                p1++;
                p2++;
            }
        }
        int[] res = new int[list.size()];
        for (int i = 0; i < list.size(); i++) {
            res[i] = list.get(i);
        }
        return res;
    }


    public int[] intersect2(int[] nums1, int[] nums2) {
        if (nums1.length == 0 || nums2.length == 0) {
            return new int[0];
        }
        Map<Integer, Integer> map = new HashMap<>();
        List<Integer> list = new ArrayList<>();
        for (int num : nums1) {
            if (map.containsKey(num)) {
                map.put(num, map.get(num) + 1);
            } else {
                map.put(num, 1);
            }
        }
        for (int num : nums2) {
            if (map.containsKey(num)) {
                map.put(num, map.get(num) - 1);
                if (map.get(num) == 0) {
                    map.remove(num);
                }
                list.add(num);
            }
        }
        int size = list.size();
        int[] res = new int[size];
        for (int i = 0; i < size; i++) {
            res[i] = list.get(i);
        }
        return res;
    }
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