LeetCode: Longest Consecutive Sequence 解题报告
Longest Consecutive Sequence
Given an unsorted array of integers, find the length of the longest consecutive elements sequence.
For example,
Given [100, 4, 200, 1, 3, 2],
The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4.
Your algorithm should run in O(n) complexity.
SOLUTION1:
用HashMap来空间换时间.
1. 在map中创建一些集合来表示连续的空间。比如,如果有[3,4,5]这样的一个集合,我们表示为key:3, value:5和key:5, value3两个集合,并且把这2个放在hashmap中。这样我们可以在O(1)的时间查询某个数字开头和结尾的集合。
2. 来了一个新的数字时,比如:N=6,我们可以搜索以N-1结尾 以N+1开头的集合有没有存在。从1中可以看到,key:5是存在的,这样我们可以删除3,5和5,3这两个key-value对,同样我们要查以7起头的集合有没有存在,同样可以删除以7起始的集合。删除后我们可以更新left,right的值,也就是合并和扩大集合。
3. 合并以上这些集合,创建一个以新的left,right作为开头,结尾的集合,分别以left, right作为key存储在map中。并且更新max (表示最长连续集合)
public class Solution {
public int longestConsecutive(int[] num) {
if (num == null) {
return 0;
} HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); int max = 0; int len = num.length;
for (int i = 0; i < len ; i++) {
// 寻找以num[i] 起头或是结尾的,如果找到,则可以跳过,因为我们
// 不需要重复的数字
if (map.get(num[i]) != null) {
continue;
} int left = num[i];
int right = num[i]; // 寻找左边界
Integer board = map.get(num[i] - 1);
if (board != null && board < left) {
// 更新左边界
left = board; // 删除左边2个集合
map.remove(left);
map.remove(num[i] - 1);
} // 寻找右边界
board = map.get(num[i] + 1);
if (board != null && board > right) {
// 更新右边界
right = board; // 删除右边2个集合
map.remove(right);
map.remove(num[i] + 1);
} // 创建新的合并之后的集合
map.put(left, right);
map.put(right, left); max = Math.max(max, right - left + 1);
} return max;
}
}
SOLUTION2:
引自大神的解法:http://blog.csdn.net/fightforyourdream/article/details/15024861
我们可以把所有的数字放在hashset中,来一个数字后,取出HashSet中的某一元素x,找x-1,x-2....x+1,x+2...是否也在set里。
// solution 2: use hashset.
public int longestConsecutive(int[] num) {
if (num == null) {
return 0;
} HashSet<Integer> set = new HashSet<Integer>();
for (int i: num) {
set.add(i);
} int max = 0;
for (int i: num) {
int cnt = 1;
set.remove(i); int tmp = i - 1;
while (set.contains(tmp)) {
set.remove(tmp);
cnt++;
tmp--;
} tmp = i + 1;
while (set.contains(tmp)) {
set.remove(tmp);
cnt++;
tmp++;
} max = Math.max(max, cnt);
} return max;
}
2015.1.2 redo:
public class Solution {
public int longestConsecutive(int[] num) {
if (num == null) {
return 0;
} HashSet<Integer> set = new HashSet<Integer>();
for (int i: num) {
set.add(i);
} int max = 0; for (int i: num) {
set.remove(i);
int sum = 1; int tmp = i - 1;
while (set.contains(tmp)) {
// bug 1:forget to add the remove statement.
set.remove(tmp);
sum++;
tmp--;
} tmp = i + 1;
while (set.contains(tmp)) {
set.remove(tmp);
sum++;
tmp++;
} max = Math.max(max, sum);
} return max;
}
}
GITHUB:
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