349. Intersection of Two Arrays

Javascript

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersection = function(nums1, nums2) {
        var rec=[];
        var result=[];
        var count=0;
        for(var i=0;i

C++和Java想得太复杂了，看了下别人的解答

优解

Java，用哈希表解决了重复的问题

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

Java，排序后用两根指针依次比对数组找出相同元素

public class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        Set 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;
    }
}

C++

class Solution {
public:
    vector intersection(vector& nums1, vector& nums2) {
    set s(nums1.begin(), nums1.end());
    vector out;
    for (int x : nums2)
        if (s.erase(x))
            out.push_back(x);
    return out;
    }
};