245 lines
7.2 KiB
Markdown
245 lines
7.2 KiB
Markdown
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<!-- GFM-TOC -->
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* [有序数组的 Two Sum](#有序数组的-two-sum)
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* [两数平方和](#两数平方和)
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* [反转字符串中的元音字符](#反转字符串中的元音字符)
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* [回文字符串](#回文字符串)
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* [归并两个有序数组](#归并两个有序数组)
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* [判断链表是否存在环](#判断链表是否存在环)
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* [最长子序列](#最长子序列)
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<!-- GFM-TOC -->
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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。
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# 有序数组的 Two Sum
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[Leetcode ：167. Two Sum II - Input array is sorted (Easy)](https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/)
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```html
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Input: numbers={2, 7, 11, 15}, target=9
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Output: index1=1, index2=2
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```
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题目描述：在有序数组中找出两个数，使它们的和为 target。
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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。
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- 如果两个指针指向元素的和 sum == target，那么得到要求的结果；
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- 如果 sum > target，移动较大的元素，使 sum 变小一些；
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- 如果 sum < target，移动较小的元素，使 sum 变大一些。
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```java
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public int[] twoSum(int[] numbers, int target) {
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int i = 0, j = numbers.length - 1;
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while (i < j) {
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int sum = numbers[i] + numbers[j];
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if (sum == target) {
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return new int[]{i + 1, j + 1};
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} else if (sum < target) {
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i++;
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} else {
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j--;
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}
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}
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return null;
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}
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```
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# 两数平方和
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[633. Sum of Square Numbers (Easy)](https://leetcode.com/problems/sum-of-square-numbers/description/)
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```html
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Input: 5
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Output: True
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Explanation: 1 * 1 + 2 * 2 = 5
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```
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题目描述：判断一个数是否为两个数的平方和。
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```java
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public boolean judgeSquareSum(int c) {
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int i = 0, j = (int) Math.sqrt(c);
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while (i <= j) {
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int powSum = i * i + j * j;
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if (powSum == c) {
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return true;
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} else if (powSum > c) {
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j--;
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} else {
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i++;
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}
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}
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return false;
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}
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```
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# 反转字符串中的元音字符
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[345. Reverse Vowels of a String (Easy)](https://leetcode.com/problems/reverse-vowels-of-a-string/description/)
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```html
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Given s = "leetcode", return "leotcede".
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```
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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。
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```java
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private final static HashSet<Character> vowels = new HashSet<>(Arrays.asList('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'));
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public String reverseVowels(String s) {
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int i = 0, j = s.length() - 1;
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char[] result = new char[s.length()];
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while (i <= j) {
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char ci = s.charAt(i);
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char cj = s.charAt(j);
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if (!vowels.contains(ci)) {
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result[i++] = ci;
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} else if (!vowels.contains(cj)) {
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result[j--] = cj;
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} else {
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result[i++] = cj;
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result[j--] = ci;
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}
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}
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return new String(result);
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}
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```
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# 回文字符串
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[680. Valid Palindrome II (Easy)](https://leetcode.com/problems/valid-palindrome-ii/description/)
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```html
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Input: "abca"
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Output: True
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Explanation: You could delete the character 'c'.
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```
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题目描述：可以删除一个字符，判断是否能构成回文字符串。
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```java
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public boolean validPalindrome(String s) {
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int i = -1, j = s.length();
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while (++i < --j) {
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if (s.charAt(i) != s.charAt(j)) {
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return isPalindrome(s, i, j - 1) || isPalindrome(s, i + 1, j);
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}
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}
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return true;
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}
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private boolean isPalindrome(String s, int i, int j) {
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while (i < j) {
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if (s.charAt(i++) != s.charAt(j--)) {
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return false;
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}
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}
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return true;
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}
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```
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# 归并两个有序数组
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[88. Merge Sorted Array (Easy)](https://leetcode.com/problems/merge-sorted-array/description/)
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```html
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Input:
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nums1 = [1,2,3,0,0,0], m = 3
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nums2 = [2,5,6], n = 3
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Output: [1,2,2,3,5,6]
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```
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题目描述：把归并结果存到第一个数组上。
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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。
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```java
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public void merge(int[] nums1, int m, int[] nums2, int n) {
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int index1 = m - 1, index2 = n - 1;
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int indexMerge = m + n - 1;
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while (index1 >= 0 || index2 >= 0) {
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if (index1 < 0) {
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nums1[indexMerge--] = nums2[index2--];
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} else if (index2 < 0) {
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nums1[indexMerge--] = nums1[index1--];
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} else if (nums1[index1] > nums2[index2]) {
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nums1[indexMerge--] = nums1[index1--];
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} else {
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nums1[indexMerge--] = nums2[index2--];
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}
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}
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}
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```
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# 判断链表是否存在环
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[141. Linked List Cycle (Easy)](https://leetcode.com/problems/linked-list-cycle/description/)
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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。
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```java
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public boolean hasCycle(ListNode head) {
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if (head == null) {
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return false;
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}
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ListNode l1 = head, l2 = head.next;
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while (l1 != null && l2 != null && l2.next != null) {
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if (l1 == l2) {
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return true;
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}
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l1 = l1.next;
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l2 = l2.next.next;
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}
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return false;
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}
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```
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# 最长子序列
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[524. Longest Word in Dictionary through Deleting (Medium)](https://leetcode.com/problems/longest-word-in-dictionary-through-deleting/description/)
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```
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Input:
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s = "abpcplea", d = ["ale","apple","monkey","plea"]
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Output:
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"apple"
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```
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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。
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```java
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public String findLongestWord(String s, List<String> d) {
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String longestWord = "";
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for (String target : d) {
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int l1 = longestWord.length(), l2 = target.length();
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if (l1 > l2 || (l1 == l2 && longestWord.compareTo(target) < 0)) {
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continue;
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}
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if (isValid(s, target)) {
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longestWord = target;
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}
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}
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return longestWord;
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}
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private boolean isValid(String s, String target) {
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int i = 0, j = 0;
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while (i < s.length() && j < target.length()) {
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if (s.charAt(i) == target.charAt(j)) {
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j++;
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}
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i++;
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}
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return j == target.length();
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}
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```
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</br><div align="center">🎨️欢迎关注我的公众号 CyC2018，在公众号后台回复关键字 **资料** 可领取复习大纲，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。可以说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>
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