LeetCode
- 关键词：LeetCode-Java
LeetCode
本文结构：
- 1.题目编号-题目标题-难度
- 2.题目内容
- 3.题目用例
- 4.解题关键点
- 5.题目答案
写在前面
Java输入输出
关键方法：next() nextInt() nextLine()
- nextInt(): it only reads the int value, nextInt() places the cursor in the same line after reading the input.
- next(): read the input only till the space. It can’t read two words separated by space. Also, next() places the cursor in the same line after reading the input.
- nextLine(): reads input including space between the words (that is, it reads till the end of line \n). Once the input is read, nextLine() positions the cursor in the next line.
Example:
输入描述:
输入包括2行：
第一行为整数n(1 <= n <= 50)，即抹除一个数之后剩下的数字个数
第二行为n个整数num[i] (1 <= num[i] <= 1000000000)
import java.util.*;
public class Next {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int[] num = new int[N];
for (int i = 0; i < N; i++) {
num[i] = sc.nextInt();
}
System.out.println("");
}
}
import java.util.*;
public class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = sc.nextLine();
char[] arr = str.toCharArray();
}
}
1. Two Sum [Medium]
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
解题思路：
[Java] 解法
public class Solution {
public int[] twoSum(int[] nums, int target) {
HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();
int[] res = new int[2];
for (int i = 0; i < nums.length; ++i) {
m.put(nums[i], i);
}
for (int i = 0; i < nums.length; ++i) {
int t = target - nums[i];
if (m.containsKey(t) && m.get(t) != i) {
res[0] = i;
res[1] = m.get(t);
break;
}
}
return res;
}
}
public class Solution {
public int[] twoSum(int[] nums, int target) {
HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();
int[] res = new int[2];
for (int i = 0; i < nums.length; ++i) {
if (m.containsKey(target - nums[i])) {
res[0] = i;
res[1] = m.get(target - nums[i]);
break;
}
m.put(nums[i], i);
}
return res;
}
}
2. Add Two Numbers [Medium]
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
解题思路：
[Java] 解法
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(-1);
ListNode cur = dummy;
int carry = 0;
while (l1 != null || l2 != null) {
int d1 = l1 == null ? 0 : l1.val;
int d2 = l2 == null ? 0 : l2.val;
int sum = d1 + d2 + carry;
carry = sum >= 10 ? 1 : 0;
cur.next = new ListNode(sum % 10);
cur = cur.next;
if (l1 != null) l1 = l1.next;
if (l2 != null) l2 = l2.next;
}
if (carry == 1) cur.next = new ListNode(1);
return dummy.next;
}
}
3. Longest Substring Without Repeating Characters [Medium]
Given a string, find the length of the longest substring without repeating characters.
Example:
Input: "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Note that the answer must be a substring, "pwke" is a _subsequence_ and not a substring.
解题思路：
[Java] 解法
public class Solution {
public int lengthOfLongestSubstring(String s) {
int[] m = new int[256];
Arrays.fill(m, -1);
int res = 0, left = -1;
for (int i = 0; i < s.length(); ++i) {
left = Math.max(left, m[s.charAt(i)]);
m[s.charAt(i)] = i;
res = Math.max(res, i - left);
}
return res;
}
}
public class Solution {
public int lengthOfLongestSubstring(String s) {
int res = 0, left = 0, right = 0;
HashSet<Character> t = new HashSet<Character>();
while (right < s.length()) {
if (!t.contains(s.charAt(right))) {
t.add(s.charAt(right++));
res = Math.max(res, t.size());
} else {
t.remove(s.charAt(left++));
}
}
return res;
}
}
4. Median of Two Sorted Arrays [Hard]
There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
You may assume nums1 and nums2 cannot be both empty.
Example:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
解题思路：
[Java] 解法
public class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int m = nums1.length, n = nums2.length, left = (m + n + 1) / 2, right = (m + n + 2) / 2;
return (findKth(nums1, 0, nums2, 0, left) + findKth(nums1, 0, nums2, 0, right)) / 2.0;
}
int findKth(int[] nums1, int i, int[] nums2, int j, int k) {
if (i >= nums1.length) return nums2[j + k - 1];
if (j >= nums2.length) return nums1[i + k - 1];
if (k == 1) return Math.min(nums1[i], nums2[j]);
int midVal1 = (i + k / 2 - 1 < nums1.length) ? nums1[i + k / 2 - 1] : Integer.MAX_VALUE;
int midVal2 = (j + k / 2 - 1 < nums2.length) ? nums2[j + k / 2 - 1] : Integer.MAX_VALUE;
if (midVal1 < midVal2) {
return findKth(nums1, i + k / 2, nums2, j, k - k / 2);
} else {
return findKth(nums1, i, nums2, j + k / 2, k - k / 2);
}
}
}
public class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int m = nums1.length, n = nums2.length, left = (m + n + 1) / 2, right = (m + n + 2) / 2;
return (findKth(nums1, nums2, left) + findKth(nums1, nums2, right)) / 2.0;
}
int findKth(int[] nums1, int[] nums2, int k) {
int m = nums1.length, n = nums2.length;
if (m == 0) return nums2[k - 1];
if (n == 0) return nums1[k - 1];
if (k == 1) return Math.min(nums1[0], nums2[0]);
int i = Math.min(m, k / 2), j = Math.min(n, k / 2);
if (nums1[i - 1] > nums2[j - 1]) {
return findKth(nums1, Arrays.copyOfRange(nums2, j, n), k - j);
} else {
return findKth(Arrays.copyOfRange(nums1, i, m), nums2, k - i);
}
}
}
public class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int m = nums1.length, n = nums2.length;
if (m < n) return findMedianSortedArrays(nums2, nums1);
if (n == 0) return (nums1[(m - 1) / 2] + nums1[m / 2]) / 2.0;
int left = 0, right = 2 * n;
while (left <= right) {
int mid2 = (left + right) / 2;
int mid1 = m + n - mid2;
double L1 = mid1 == 0 ? Double.MIN_VALUE : nums1[(mid1 - 1) / 2];
double L2 = mid2 == 0 ? Double.MIN_VALUE : nums2[(mid2 - 1) / 2];
double R1 = mid1 == m * 2 ? Double.MAX_VALUE : nums1[mid1 / 2];
double R2 = mid2 == n * 2 ? Double.MAX_VALUE : nums2[mid2 / 2];
if (L1 > R2) left = mid2 + 1;
else if (L2 > R1) right = mid2 - 1;
else return (Math.max(L1, L2) + Math.min(R1, R2)) / 2;
}
return -1;
}
}
5. Longest Palindromic Substring [Medium]
Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
Example:
Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.
Input: "cbbd"
Output: "bb"
解题思路：
[C++] 解法
class Solution {
public:
string longestPalindrome(string s) {
if (s.empty()) return "";
int n = s.size(), dp[n][n] = {0}, left = 0, len = 1;
for (int i = 0; i < n; ++i) {
dp[i][i] = 1;
for (int j = 0; j < i; ++j) {
dp[j][i] = (s[i] == s[j] && (i - j < 2 || dp[j + 1][i - 1]));
if (dp[j][i] && len < i - j + 1) {
len = i - j + 1;
left = j;
}
}
}
return s.substr(left, len);
}
};
6. ZigZag Conversion [Easy]
The string “PAYPALISHIRING” is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
P A H N
A P L S I I G
Y I R
And then read line by line: “PAHNAPLSIIGYIR”
Write the code that will take a string and make this conversion given a number of rows:
string convert(string s, int numRows);
Example:
Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"
Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:
P I N
A L S I G
Y A H R
P I
解题思路：
[C++] 解法
class Solution {
public:
string convert(string s, int numRows) {
if (numRows <= 1) return s;
string res;
int i = 0, n = s.size();
vector<string> vec(numRows);
while (i < n) {
for (int pos = 0; pos < numRows && i < n; ++pos) {
vec[pos] += s[i++];
}
for (int pos = numRows - 2; pos >= 1 && i < n; --pos) {
vec[pos] += s[i++];
}
}
for (auto &a : vec) res += a;
return res;
}
};
7. Reverse Integer [Easy]
Given a 32-bit signed integer, reverse digits of an integer.
Example:
Input: 123
Output: 321
Input: -123
Output: -321
Input: 120
Output: 21
Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
解题思路：
[C++] 解法
class Solution {
public:
int reverse(int x) {
long res = 0;
while (x != 0) {
res = 10 * res + x % 10;
x /= 10;
}
return (res > INT_MAX || res < INT_MIN) ? 0 : res;
}
};
8. String to Integer [Easy]
The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.
The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.
If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.
If no valid conversion could be performed, a zero value is returned.
Example:
Input: "42"
Output: 42
Input: " -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
Then take as many numerical digits as possible, which gets 42.
Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.
Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical
digit or a +/- sign. Therefore no valid conversion could be performed.
Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
Thefore INT_MIN (−231) is returned.
解题思路：
[Java] 解法
public class Solution {
public int myAtoi(String str) {
if (str.isEmpty()) return 0;
int sign = 1, base = 0, i = 0, n = str.length();
while (i < n && str.charAt(i) == ' ') ++i;
if (i < n && (str.charAt(i) == '+' || str.charAt(i) == '-')) {
sign = (str.charAt(i++) == '+') ? 1 : -1;
}
while (i < n && str.charAt(i) >= '0' && str.charAt(i) <= '9') {
if (base > Integer.MAX_VALUE / 10 || (base == Integer.MAX_VALUE / 10 && str.charAt(i) - '0' > 7)) {
return (sign == 1) ? Integer.MAX_VALUE : Integer.MIN_VALUE;
}
base = 10 * base + (str.charAt(i++) - '0');
}
return base * sign;
}
}
9. Palindrome Number [Easy]
Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.
Example:
Input: 121
Output: true
Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
Coud you solve it without converting the integer to a string?
解题思路：
[C++] 解法
class Solution {
public:
bool isPalindrome(int x) {
if (x < 0) return false;
int div = 1;
while (x / div >= 10) div *= 10;
while (x > 0) {
int left = x / div;
int right = x % 10;
if (left != right) return false;
x = (x % div) / 10;
div /= 100;
}
return true;
}
};
class Solution {
public:
bool isPalindrome(int x) {
if (x < 0 || (x % 10 == 0 && x != 0)) return false;
int revertNum = 0;
while (x > revertNum) {
revertNum = revertNum * 10 + x % 10;
x /= 10;
}
return x == revertNum || x == revertNum / 10;
}
};
class Solution {
public:
bool isPalindrome(int x) {
if (x < 0 || (x % 10 == 0 && x != 0)) return false;
return reverse(x) == x;
}
int reverse(int x) {
int res = 0;
while (x != 0) {
if (res > INT_MAX / 10) return -1;
res = res * 10 + x % 10;
x /= 10;
}
return res;
}
};
10. Regular Expression Matching [Hard]
Given an input string (s) and a pattern (p), implement regular expression matching with support for ‘.’ and ‘*’.
'.' Matches any single character.
'*' Matches zero or more of the preceding element.
Note:
s could be empty and contains only lowercase letters a-z.
p could be empty and contains only lowercase letters a-z, and characters like . or *.
Example:
Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".
Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".
Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".
Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".
Input:
s = "mississippi"
p = "mis*is*p*."
Output: false
解题思路：
[C++] 解法
class Solution {
public:
bool isMatch(string s, string p) {
if (p.empty()) return s.empty();
if (p.size() > 1 && p[1] == '*') {
return isMatch(s, p.substr(2)) || (!s.empty() && (s[0] == p[0] || p[0] == '.') && isMatch(s.substr(1), p));
} else {
return !s.empty() && (s[0] == p[0] || p[0] == '.') && isMatch(s.substr(1), p.substr(1));
}
}
};
参考资料
https://blog.csdn.net/weixin_37605770/article/details/79749347
https://github.com/grandyang/leetcode