CS-Notes/docs/notes/Leetcode 题解 - 双指针.md

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* [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)
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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
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
- 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
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```java
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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;
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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
Output: True
Explanation: 1 * 1 + 2 * 2 = 5
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```
题目描述:判断一个数是否为两个数的平方和。
```java
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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;
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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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```
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
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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);
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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"
Output: True
Explanation: You could delete the character 'c'.
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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。
```java
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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;
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}
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private boolean isPalindrome(String s, int i, int j) {
while (i < j) {
if (s.charAt(i++) != s.charAt(j--)) {
return false;
}
}
return true;
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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
Input:
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nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6], n = 3
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Output: [1,2,2,3,5,6]
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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) {
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--];
}
}
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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。
```java
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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;
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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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```
Input:
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s = "abpcplea", d = ["ale","apple","monkey","plea"]
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Output:
"apple"
```
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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) {
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;
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}
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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();
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}
```
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