求数组交集不同解法小结
求数组交集不同解法小结
- 求数组交集不同解法小结
- 声明
- LintCode547求数组交集_要求元素不重复
- 解法一排序二分查找
- 解法二HasSet暴力去重
- 解法三双指针法重视
- LintCode548求数组交集变种
- 解法一HashMap统计次数实现
- 解法二排序二分查找变种双指针
声明
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/
LintCode547：求数组交集_要求元素不重复
LintCode547，给出两个数组，求二者交集且元素不重复， O(N2) 查找会超时；
解法一：排序+二分查找
O(N2) 算法超时主要发生在大数组查找过程，因此采用二分查找提升查找效率，交集用HashSet保存实现去重；
/**
* 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 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；
/**
* 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 hash1 = new HashSet<>();
for (int i = 0; i < num1.length; i++) {
hash1.add(num1[i]);
}
HashSet 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）
通过双指针求交集，必须首先将求交集的两数组排序；
/**
* 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；
解法一：HashMap统计次数实现
通过HashMap记录数组中每个元素与对应的出现次数；
/**
* 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 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 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：与经典二分不同，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：经过排序后，假设两数组中拥有某个交集元素cur, 通过二分查找到cur在第二个数组中的位置index,通过双指针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；
/**
* 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 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;
}
}