题目

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.

思路

• （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

代码

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 usedfor (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;}}
}