Difficulty: 简单
编写一个函数来查找字符串数组中的最长公共前缀。
如果不存在公共前缀,返回空字符串 ""
。
示例 1:
输入:strs = ["flower","flow","flight"]
输出:"fl"
示例 2:
输入:strs = ["dog","racecar","car"]
输出:""
解释:输入不存在公共前缀。
提示:
0 <= strs.length <= 200
0 <= strs[i].length <= 200
strs[i]
仅由小写英文字母组成
**思路1:横向比较。**从到到尾遍历字符串数组,找出strs[0]和strs[1]的最长公共前缀prefix,然后再让prefix与strs[2]比较...,以此类推。时间复杂度:O(m*n),其中 m 是字符串数组中的字符串的平均长度,n是字符串的数量。
**思路2:纵向比较。**取第一个字符串的第0个字符,看是否所有字符串的第0个都是这个字符;然后再取第1个,第二个...,以此类推。时间复杂度:O(m*n)
**思路3:分治法。**显然这个问题和归并排序一样,可以先求一部分的,再融合起来。根据Master公式,时间复杂度位O(m*n)
//方法1:横向比较
public String longestCommonPrefix(String[] strs) {
if (strs == null || strs.length == 0) {
return "";
}
String prefix = strs[0];
int count = strs.length;
for (int i = 1; i < count; i++) {
prefix = longestCommonPrefix(prefix, strs[i]);
if (prefix.length() == 0) {
break;
}
}
return prefix;
}
public String longestCommonPrefix(String str1, String str2) {
int i = 0;
while (i < str1.length() && i < str2.length() && str1.charAt(i) == str2.charAt(i)) {
i++;
}
return str1.substring(0, i);
}
//方法二:纵向比较
public String longestCommonPrefix(String[] strs) {
if (strs == null || strs.length == 0) {
return "";
}
for (int i = 0; i < strs[0].length(); i++) {
char c = strs[0].charAt(i);
for (int j = 1; j < strs.length; j++) {
if (i == strs[j].length() || strs[j].charAt(i) != c) {
return strs[0].substring(0, i);
}
}
}
return strs[0];
}
//方法三:分治法
public String longestCommonPrefix(String[] strs) {
if (strs == null || strs.length == 0) {
return "";
} else {
return longestCommonPrefix(strs, 0, strs.length - 1);
}
}
public String longestCommonPrefix(String[] strs, int start, int end) {
if (start == end) {
return strs[start];
} else {
int mid = (end - start) / 2 + start;
//求一下左右子数组结果的longestCommonPrefix即可。
String lcpLeft = longestCommonPrefix(strs, start, mid);
String lcpRight = longestCommonPrefix(strs, mid + 1, end);
return longestCommonPrefix(lcpLeft, lcpRight);
}
}
public String longestCommonPrefix(String str1, String str2) {
int i = 0;
while (i < str1.length() && i < str2.length() && str1.charAt(i) == str2.charAt(i)) {
i++;
}
return str1.substring(0, i);
}
Difficulty: 困难
给你一个链表数组,每个链表都已经按升序排列。
请你将所有链表合并到一个升序链表中,返回合并后的链表。
示例 1:
输入:lists = [[1,4,5],[1,3,4],[2,6]]
输出:[1,1,2,3,4,4,5,6]
解释:链表数组如下:
[
1->4->5,
1->3->4,
2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6
示例 2:
输入:lists = []
输出:[]
示例 3:
输入:lists = [[]]
输出:[]
提示:
k == lists.length
0 <= k <= 10^4
0 <= lists[i].length <= 500
-10^4 <= lists[i][j] <= 10^4
lists[i]
按 升序 排列lists[i].length
的总和不超过10^4
归并:
public ListNode mergeKLists(ListNode[] lists) {
if (lists == null || lists.length == 0) {
return null;
}
return mergeKLists(lists, 0, lists.length - 1);
}
private ListNode mergeKLists(ListNode[] lists, int left, int right) {
if (left == right) {
return lists[left];
}
int mid = left + (right - left) / 2; //int mid = (left+right)>>>1;
return merge(mergeKLists(lists, left, mid), mergeKLists(lists, mid + 1, right));
}
private ListNode merge(ListNode l1, ListNode l2) {
if (l1 == null || l2 == null) {
return l1 == null ? l2 : l1;
}
if (l1.val < l2.val) {
l1.next = merge(l1.next, l2);
return l1;
} else {
l2.next = merge(l1, l2.next);
return l2;
}
}
PQ:
public ListNode mergeKLists(ListNode[] lists) {
if (lists == null || lists.length == 0) return null;
PriorityQueue<ListNode> queue = new PriorityQueue<>(lists.length, new Comparator<ListNode>() {
@Override
public int compare(ListNode o1, ListNode o2) {
return o1.val - o2.val;
}
});
ListNode dummy = new ListNode(0);
ListNode p = dummy;
for (ListNode node : lists) {
if (node != null) queue.add(node);
}
while (!queue.isEmpty()) {
p.next = queue.poll();
p = p.next;
if (p.next != null) queue.add(p.next);
}
return dummy.next;
}
//或者用lambda表达式
public ListNode mergeKLists(ListNode[] lists) {
if (lists == null || lists.length == 0) return null;
PriorityQueue<ListNode> queue = new PriorityQueue<>(lists.length,
(a,b)-> (a.val - b.val)
);
ListNode dummy = new ListNode(0);
ListNode p = dummy;
for (ListNode node : lists) {
if (node != null) queue.add(node);
}
while (!queue.isEmpty()) {
p.next = queue.poll();
p = p.next;
if (p.next != null) queue.add(p.next);
}
return dummy.next;
}