245 lines
7.2 KiB
Markdown
245 lines
7.2 KiB
Markdown
<!-- GFM-TOC -->
|
||
* [有序数组的 Two Sum](#有序数组的-two-sum)
|
||
* [两数平方和](#两数平方和)
|
||
* [反转字符串中的元音字符](#反转字符串中的元音字符)
|
||
* [回文字符串](#回文字符串)
|
||
* [归并两个有序数组](#归并两个有序数组)
|
||
* [判断链表是否存在环](#判断链表是否存在环)
|
||
* [最长子序列](#最长子序列)
|
||
<!-- GFM-TOC -->
|
||
|
||
|
||
双指针主要用于遍历数组，两个指针指向不同的元素，从而协同完成任务。
|
||
|
||
# 有序数组的 Two Sum
|
||
|
||
[Leetcode ：167. Two Sum II - Input array is sorted (Easy)](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/)
|
||
|
||
```html
|
||
Input: numbers={2, 7, 11, 15}, target=9
|
||
Output: index1=1, index2=2
|
||
```
|
||
|
||
题目描述：在有序数组中找出两个数，使它们的和为 target。
|
||
|
||
使用双指针，一个指针指向值较小的元素，一个指针指向值较大的元素。指向较小元素的指针从头向尾遍历，指向较大元素的指针从尾向头遍历。
|
||
|
||
- 如果两个指针指向元素的和 sum == target，那么得到要求的结果；
|
||
- 如果 sum > target，移动较大的元素，使 sum 变小一些；
|
||
- 如果 sum < target，移动较小的元素，使 sum 变大一些。
|
||
|
||
```java
|
||
public int[] twoSum(int[] numbers, int target) {
|
||
int i = 0, j = numbers.length - 1;
|
||
while (i < j) {
|
||
int sum = numbers[i] + numbers[j];
|
||
if (sum == target) {
|
||
return new int[]{i + 1, j + 1};
|
||
} else if (sum < target) {
|
||
i++;
|
||
} else {
|
||
j--;
|
||
}
|
||
}
|
||
return null;
|
||
}
|
||
```
|
||
|
||
# 两数平方和
|
||
|
||
[633. Sum of Square Numbers (Easy)](https://leetcode.com/problems/sum-of-square-numbers/description/)
|
||
|
||
```html
|
||
Input: 5
|
||
Output: True
|
||
Explanation: 1 * 1 + 2 * 2 = 5
|
||
```
|
||
|
||
题目描述：判断一个数是否为两个数的平方和。
|
||
|
||
```java
|
||
public boolean judgeSquareSum(int c) {
|
||
int i = 0, j = (int) Math.sqrt(c);
|
||
while (i <= j) {
|
||
int powSum = i * i + j * j;
|
||
if (powSum == c) {
|
||
return true;
|
||
} else if (powSum > c) {
|
||
j--;
|
||
} else {
|
||
i++;
|
||
}
|
||
}
|
||
return false;
|
||
}
|
||
```
|
||
|
||
# 反转字符串中的元音字符
|
||
|
||
[345. Reverse Vowels of a String (Easy)](https://leetcode.com/problems/reverse-vowels-of-a-string/description/)
|
||
|
||
```html
|
||
Given s = "leetcode", return "leotcede".
|
||
```
|
||
|
||
使用双指针指向待反转的两个元音字符，一个指针从头向尾遍历，一个指针从尾到头遍历。
|
||
|
||
```java
|
||
private final static HashSet<Character> vowels = new HashSet<>(Arrays.asList('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'));
|
||
|
||
public String reverseVowels(String s) {
|
||
int i = 0, j = s.length() - 1;
|
||
char[] result = new char[s.length()];
|
||
while (i <= j) {
|
||
char ci = s.charAt(i);
|
||
char cj = s.charAt(j);
|
||
if (!vowels.contains(ci)) {
|
||
result[i++] = ci;
|
||
} else if (!vowels.contains(cj)) {
|
||
result[j--] = cj;
|
||
} else {
|
||
result[i++] = cj;
|
||
result[j--] = ci;
|
||
}
|
||
}
|
||
return new String(result);
|
||
}
|
||
```
|
||
|
||
# 回文字符串
|
||
|
||
[680. Valid Palindrome II (Easy)](https://leetcode.com/problems/valid-palindrome-ii/description/)
|
||
|
||
```html
|
||
Input: "abca"
|
||
Output: True
|
||
Explanation: You could delete the character 'c'.
|
||
```
|
||
|
||
题目描述：可以删除一个字符，判断是否能构成回文字符串。
|
||
|
||
```java
|
||
public boolean validPalindrome(String s) {
|
||
int i = -1, j = s.length();
|
||
while (++i < --j) {
|
||
if (s.charAt(i) != s.charAt(j)) {
|
||
return isPalindrome(s, i, j - 1) || isPalindrome(s, i + 1, j);
|
||
}
|
||
}
|
||
return true;
|
||
}
|
||
|
||
private boolean isPalindrome(String s, int i, int j) {
|
||
while (i < j) {
|
||
if (s.charAt(i++) != s.charAt(j--)) {
|
||
return false;
|
||
}
|
||
}
|
||
return true;
|
||
}
|
||
```
|
||
|
||
# 归并两个有序数组
|
||
|
||
[88. Merge Sorted Array (Easy)](https://leetcode.com/problems/merge-sorted-array/description/)
|
||
|
||
```html
|
||
Input:
|
||
nums1 = [1,2,3,0,0,0], m = 3
|
||
nums2 = [2,5,6], n = 3
|
||
|
||
Output: [1,2,2,3,5,6]
|
||
```
|
||
|
||
题目描述：把归并结果存到第一个数组上。
|
||
|
||
需要从尾开始遍历，否则在 nums1 上归并得到的值会覆盖还未进行归并比较的值。
|
||
|
||
```java
|
||
public void merge(int[] nums1, int m, int[] nums2, int n) {
|
||
int index1 = m - 1, index2 = n - 1;
|
||
int indexMerge = m + n - 1;
|
||
while (index1 >= 0 || index2 >= 0) {
|
||
if (index1 < 0) {
|
||
nums1[indexMerge--] = nums2[index2--];
|
||
} else if (index2 < 0) {
|
||
nums1[indexMerge--] = nums1[index1--];
|
||
} else if (nums1[index1] > nums2[index2]) {
|
||
nums1[indexMerge--] = nums1[index1--];
|
||
} else {
|
||
nums1[indexMerge--] = nums2[index2--];
|
||
}
|
||
}
|
||
}
|
||
```
|
||
|
||
# 判断链表是否存在环
|
||
|
||
[141. Linked List Cycle (Easy)](https://leetcode.com/problems/linked-list-cycle/description/)
|
||
|
||
使用双指针，一个指针每次移动一个节点，一个指针每次移动两个节点，如果存在环，那么这两个指针一定会相遇。
|
||
|
||
```java
|
||
public boolean hasCycle(ListNode head) {
|
||
if (head == null) {
|
||
return false;
|
||
}
|
||
ListNode l1 = head, l2 = head.next;
|
||
while (l1 != null && l2 != null && l2.next != null) {
|
||
if (l1 == l2) {
|
||
return true;
|
||
}
|
||
l1 = l1.next;
|
||
l2 = l2.next.next;
|
||
}
|
||
return false;
|
||
}
|
||
```
|
||
|
||
# 最长子序列
|
||
|
||
[524. Longest Word in Dictionary through Deleting (Medium)](https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/description/)
|
||
|
||
```
|
||
Input:
|
||
s = "abpcplea", d = ["ale","apple","monkey","plea"]
|
||
|
||
Output:
|
||
"apple"
|
||
```
|
||
|
||
题目描述：删除 s 中的一些字符，使得它构成字符串列表 d 中的一个字符串，找出能构成的最长字符串。如果有多个相同长度的结果，返回字典序的最小字符串。
|
||
|
||
```java
|
||
public String findLongestWord(String s, List<String> d) {
|
||
String longestWord = "";
|
||
for (String target : d) {
|
||
int l1 = longestWord.length(), l2 = target.length();
|
||
if (l1 > l2 || (l1 == l2 && longestWord.compareTo(target) < 0)) {
|
||
continue;
|
||
}
|
||
if (isValid(s, target)) {
|
||
longestWord = target;
|
||
}
|
||
}
|
||
return longestWord;
|
||
}
|
||
|
||
private boolean isValid(String s, String target) {
|
||
int i = 0, j = 0;
|
||
while (i < s.length() && j < target.length()) {
|
||
if (s.charAt(i) == target.charAt(j)) {
|
||
j++;
|
||
}
|
||
i++;
|
||
}
|
||
return j == target.length();
|
||
}
|
||
```
|
||
|
||
|
||
|
||
|
||
</br><div align="center">🎨 </br></br> 更多精彩内容将发布在公众号 **CyC2018**，在公众号后台回复关键字 **资料** 可领取一份技术面试复习思维导图，这份大纲是我花了一整年时间整理的面试知识点列表，不仅系统整理了面试知识点，而且标注了各个知识点的重要程度，从而帮你理清多而杂的面试知识点。我基本是按照这份大纲来进行复习的，这份大纲对我拿到了 BAT 头条等 Offer 起到很大的帮助。你们完全可以和我一样根据大纲上列的知识点来进行复习，就不用看很多不重要的内容，也可以知道哪些内容很重要从而多安排一些复习时间。</div></br>
|
||
<div align="center"><img width="180px" src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/%E5%85%AC%E4%BC%97%E5%8F%B7.jpg"></img></div>
|