标签:io os ar 使用 java for sp on cti
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.
(预处理)保存一个哈希表(或者集合),用于o(1)时间查找该数字是否存在于数组当中
- 数字有可能重复,所以其实哈希集合就够了,典型的空间换时间
(预处理)一个表征是否使用的used表,用于o(1)时间查找该数字是否已经被包含在另外一个序列当中
- 注意数字有可能是负数,所以直接用数组不好
轮询数组,遇到一个数字,查找其左,右的最长连续序列长度,并记录与已知最大长度相比较
public class Solution {
public int longestConsecutive(int[] num) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for (int i = 0; i < num.length; i++) {
map.put(num[i], i);
}
Map<Integer, Boolean> used = new HashMap<Integer, Boolean>();
for (int i : num) {
used.put(i, false);
}
int max = Integer.MIN_VALUE;
for (int i = 0; i < num.length; i++) {
if (!used.get(num[i])) {
used.put(num[i], true);
int k = num[i];
int leftLength = findLength(k, map, "left");
int rightLength = findLength(k, map, "right");
// mark used
for (int j = 0; j < leftLength; j++) {
used.put(k - j - 1, true);
}
for (int j = 0; j < rightLength; j++) {
used.put(k + j + 1, true);
}
int total = leftLength + rightLength + 1;
if (total > max) {
max = total;
}
}
}
return max;
}
private int findLength(int k, Map<Integer, Integer> map, String direction) {
if ("left".equals(direction)) {
int l = k - 1;
while (map.containsKey(l)) {
l -= 1;
}
return k - l - 1;
} else {
int l = k + 1;
while (map.containsKey(l)) {
l += 1;
}
return l - k - 1;
}
}
}[LeetCode] Longest Consecutive Sequence 求解
标签:io os ar 使用 java for sp on cti
原文地址:http://my.oschina.net/zuoyc/blog/338719