LintCode547/548_求数组交集不一样解法小结

时间 2019-11-16

标签 lintcode547 lintcode 数组交集不一样解法小结繁體版

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LintCode547/548_求数组交集不一样解法小结

[TOC]java

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文章均为本人技术笔记，转载请注明出处：
[1] https://segmentfault.com/u/yzwall
[2] blog.csdn.net/j_dark/算法

LintCode547：求数组交集_要求元素不重复

LintCode547，给出两个数组，求两者交集且元素不重复，$O(N^{2})$查找会超时；segmentfault

解法一：排序+二分查找

$O(N ^{2})$算法超时主要发生在大数组查找过程，所以采用二分查找提高查找效率，交集用HashSet保存实现去重；数组

/**
 * 解法1：排序+二分+HashSet去重
 * http://www.lintcode.com/zh-cn/problem/intersection-of-two-arrays/
 * 求数组交集，要求元素不重复出现
 * @author yzwall
 */
class Solution {
    public int[] intersection(int[] num1, int[] num2) {
        int[] results;
        if (num1 == null || num1.length == 0 || num2 == null || num2.length == 0) {
            results = new int[0];
            return results;
        }
        
        HashSet<Integer> set = new HashSet<>();
        Arrays.sort(num1);
        Arrays.sort(num2);
        int index2 = 0;
        for (int i = 0; i < num1.length; i++) {
            // num2是子集
            if (index2 > num2.length - 1) {
                break;
            }
            int index = binarySearch(num2, index2, num1[i]);
            if (index != -1) {
                // set去重
                set.add(num1[i]);
                // num2指针移动
                index2 = index;
            }
        }
        
        results = new int[set.size()];
        int i = 0;
        for (Integer cur : set) {
            results[i++] = cur.intValue();
        }
        return results;
    }
    
    // Index2~num.length - 1，经典二分查找
    private int binarySearch(int[] num, int index2, int target) {
        int start = index2;
        int end = num.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (num[mid] == target) {
                return mid;
            } else if (num[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (num[start] == target) {
            return start;
        }
        if (num[end] == target) {
            return end;
        }
        return -1;
    }
}

解法二：HasSet暴力去重

直接运用两个HashSet实现去重求交集，与解法一相比实现简单；.net

/**
 * 解法2：HashSet暴力去重
 * http://www.lintcode.com/zh-cn/problem/intersection-of-two-arrays/
 * 求数组交集，要求元素不重复出现
 * @author yzwall
 */
class Solution {
    public int[] intersection(int[] num1, int[] num2) {
        int[] results;
        if (num1 == null || num1.length == 0 || num2 == null || num2.length == 0) {
            results = new int[0];
            return results;
        }
        HashSet<Integer> hash1 = new HashSet<>();
        for (int i = 0; i < num1.length; i++) {
            hash1.add(num1[i]);
        }
        HashSet<Integer> hash2 = new HashSet<>();
        for (int i = 0; i < num2.length; i++) {
            if (hash1.contains(num2[i])) {
                hash2.add(num2[i]);
            }
        }
        
        results = new int[hash2.size()];
        int i = 0;
        for (Integer num : hash2) {
            results[i++] = num;
        }
        
        return results;
    }
}

解法三：双指针法（重视）

经过双指针求交集，必须首先将求交集的两数组排序；指针

/**
 * 解法3：双指针法
 * http://www.lintcode.com/zh-cn/problem/intersection-of-two-arrays/
 * 求数组交集，要求元素不重复出现
 * @author yzwall
 */
class Solution {
    public int[] intersection(int[] num1, int[] num2) {
        int[] results;
        if (num1 == null || num1.length == 0 || num2 == null || num2.length == 0) {
            results = new int[0];
            return results;
        }
        Arrays.sort(num1);
        Arrays.sort(num2);
        int i = 0, j = 0;
        int index = 0;
        int[] temp = new int[num1.length];
        while (i < num1.length && j < num2.length) {
            if (num1[i] == num2[j]) {
                // temp[index - 1] != num1[i]去重
                if (index == 0 || temp[index - 1] != num1[i]) {
                    temp[index++] = num1[i];
                }
                i++;
                j++;
            } else if (num1[i] < num2[j]) {
                i++;
            } else {
                j++;
            }
        }
        
        i = 0;
        results = new int[index];
        for (i = 0; i < index; i++) {
            results[i] = temp[i];
        }
        return results;
    }
}

LintCode548：求数组交集变种

在求数组交集的基础上，要求交集元素出现次数与在数组中出现次数相同；code

解法一：HashMap统计次数实现

经过HashMap<Integer, Integer>记录数组中每一个元素与对应的出现次数；blog

/**
 * 解法2：HashMap统计重复出现次数
 * http://www.lintcode.com/zh-cn/problem/intersection-of-two-arrays-ii/
 * 求两数组交集，要求交集元素按照最小出现次数出现
 * @author yzwall
 */
class Solution {
    public int[] intersection(int[] num1, int[] num2) {
        int[] results;
        if (num1 == null || num1.length == 0 || num2 == null || num2.length == 0) {
            results = new int[0];
            return results;
        }
        
        HashMap<Integer, Integer> hash = new HashMap<>();
        for (int i = 0; i < num1.length; i++) {
            if (hash.containsKey(num1[i])) {
                hash.put(num1[i], hash.get(num1[i]) + 1);
            } else {
                hash.put(num1[i], 1);
            }
        }
        
        ArrayList<Integer> list = new ArrayList<>();
        for (int i = 0; i < num2.length; i++) {
            if (hash.containsKey(num2[i]) && hash.get(num2[i]) > 0) {
                list.add(num2[i]);
                hash.put(num2[i], hash.get(num2[i]) - 1);
            }
        }
        
        results = new int[list.size()];
        for (int i = 0; i < list.size(); i++) {
            results[i] = list.get(i);
        }
        return results;
    }
}

解法二：排序+二分查找变种+双指针

变种二分查找：与经典二分不一样，解法二中二分查找用于找到查找目标第一次出现位置；
双指针解法：通过排序后，假设两数组中拥有某个交集元素cur, 经过二分查找到cur在第二个数组中的位置index,经过双指针cnt1与cnt2统计交集元素cur在两个数组中各自出现的总次数，较小者表示该交集元素在交集中出现的次数；排序

/**
 * 解法1：排序+二分查找+双指针
 * http://www.lintcode.com/zh-cn/problem/intersection-of-two-arrays-ii/
 * 求两数组交集，要求交集元素按照最小出现次数出现
 * @author yzwall
 */
class Solution3 {
    public int[] intersection(int[] num1, int[] num2) {
        int[] results;
        if (num1 == null || num1.length == 0 || num2 == null || num2.length == 0) {
            results = new int[0];
            return results;
        }
        
        ArrayList<Integer> list = new ArrayList<>();
        Arrays.sort(num1);
        Arrays.sort(num2);
        int index2 = 0;
        int i = 0;
        while(i < num1.length) {
            // num2是子集
            if (index2 > num2.length - 1) {
                break;
            }
            int cnt1 = 1, cnt2 = 1;
            int cur = num1[i];
            int index = binarySearch(num2, index2, cur);
            if (index != -1) {
                // 查找交集元素cur在数组num1中出现总次数
                for (int k = 1; k < num1.length && i + k < num1.length; k++) {
                    if (num1[i + k] != cur) {
                        break;
                    }
                    cnt1++;
                }
                // 查找交集元素cur在数组num2中出现总次数
                for (int k = 1; k < num2.length && index + k < num2.length; k++) {
                    if (num2[index + k] != cur) {
                        break;
                    }
                    cnt2++;
                }
                int min = Math.min(cnt1, cnt2);
                for (int k = 0; k < min; k++) {
                    list.add(cur);
                }
                // num2指针移动
                index2 += cnt2;
            }
            // num1指针移动
            i += cnt1;
        }
        
        results = new int[list.size()];
        i = 0;
        for (Integer cur : list) {
            results[i++] = cur.intValue();
        }
        return results;
    }
    
    // 返回target第一次出现位置，target不存在返回-1
    private int binarySearch(int[] num, int index2, int target) {
        int start = index2;
        int end = num.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (num[mid] == target) {
                end = mid;
            } else if (num[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (num[start] == target) {
            return start;
        }
        if (num[end] == target) {
            return end;
        }
        return -1;
    }
}