leetcode 简单题合集（Java版）

326. 3的幂

https://leetcode-cn.com/problems/power-of-three/

题解：

class Solution {public boolean isPowerOfThree(int n) {if (n == 0) return false;while (n != 1) {if (n % 3 != 0) return false;else n /= 3;}return true;}
}

327. 4的幂

https://leetcode-cn.com/problems/power-of-four/

题解：

class Solution {public boolean isPowerOfFour(int n) {if (n == 0) return false;while (n != 1) {if (n % 4 != 0) return false;else n /= 4;}return true;}
}

344. 反转字符串

https://leetcode-cn.com/problems/reverse-string/

class Solution {public void reverseString(char[] s) {for (int i = 0; i < s.length / 2; i++) {char t = s[i];s[i] = s[s.length - i - 1];s[s.length - i - 1] = t;}}
}

345. 反转字符串中的元音字母

https://leetcode-cn.com/problems/reverse-vowels-of-a-string/

class Solution {public static void main(String[] args) {reverseVowels("hello");}public static String reverseVowels(String s) {char[] arr = s.toCharArray();int left = 0, right = arr.length - 1;while (left < right) {while (left < arr.length && !isVowel(arr[left])) {left++;}while (right >= 0 && !isVowel(arr[right])) {right--;}if (left >= right) break;char temp = arr[left];arr[left] = arr[right];arr[right] = temp;left++;right--;}StringBuilder sb = new StringBuilder();for (int i = 0; i < arr.length; i++) {sb.append(arr[i]);}return sb.toString();}public static boolean isVowel(char c) {return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U');}
}

349. 两个数组的交集

https://leetcode-cn.com/problems/intersection-of-two-arrays/

public class Solution {public int[] intersection(int[] nums1, int[] nums2) {HashSet<Integer> set = new HashSet<>();ArrayList<Integer> list = new ArrayList<>();for (int i = 0; i < nums1.length; i++) {set.add(nums1[i]);}for (int i = 0; i < nums2.length; i++) {if (set.remove(nums2[i])) {list.add(nums2[i]);}}int[] res = new int[list.size()];for (int i = 0; i < list.size(); i++) {res[i] = list.get(i);}return res;}
}

350. 两个数组的交集 II

https://leetcode-cn.com/problems/intersection-of-two-arrays-ii/comments/

思路：HashMap存原始个数，遇到相同就减1并输出

class Solution {public int[] intersect(int[] nums1, int[] nums2) {HashMap<Integer, Integer> map = new HashMap();ArrayList<Integer> list = new ArrayList<>();for (int i = 0; i < nums1.length; i++) {Integer cnt = map.get(nums1[i]);cnt = cnt == null ? 0 : cnt;map.put(nums1[i], cnt + 1);}for (int i = 0; i < nums2.length; i++) {if (map.containsKey(nums2[i])) {int cnt = map.get(nums2[i]);if (cnt > 1) {map.put(nums2[i], cnt - 1);} else {map.remove(nums2[i]);}list.add(nums2[i]);}}int[] res = new int[list.size()];for (int i = 0; i < list.size(); i++) {res[i] = list.get(i);}return res;}
}

367. 有效的完全平方数

https://leetcode-cn.com/problems/valid-perfect-square/

二分查找，中规中矩

class Solution {public boolean isPerfectSquare(int num) {int min = 0, max = 46340;return binSearch(min, max, num);}public boolean binSearch(int min, int max, int num) {int mid = (min + max) / 2;if (min * min == num || mid * mid == num || max * max == num) {return true;}if (min >= max) return false;if (num < mid * mid) {return binSearch(min, mid, num);} else {return binSearch(mid + 1, max, num);}}
}

374. 猜数字大小

https://leetcode-cn.com/problems/guess-number-higher-or-lower/

本题是经典的二分查找题

注意 int mid = begin +(end - begin)/2; 不要用int mid = (end + begin)/2; 会越界

public class Solution extends GuessGame {public int guessNumber(int n) {return binSearch(1, n);}public int binSearch(int left, int right) {int mid = left + (right - left) / 2;int flag = guess(mid);if (flag == 0) return mid;else if (flag < 0) return binSearch(left, mid - 1);else return binSearch(mid + 1, right);}
}

383. 赎金信

https://leetcode-cn.com/problems/ransom-note/

解法1：适用于两个字串所包含字符不受任何限制的情况

class Solution {public boolean canConstruct(String ransomNote, String magazine) {char[] c1 = ransomNote.toCharArray();char[] c2 = magazine.toCharArray();HashMap<Character, Integer> map = new HashMap<>();for (int i = 0; i < c2.length; i++) {Integer cnt = map.get(c2[i]);cnt = cnt == null ? 0 : cnt;map.put(c2[i], cnt + 1);}for (int i = 0; i < c1.length; i++) {if (map.containsKey(c1[i])) {int cnt = map.get(c1[i]);if (cnt > 1) {map.put(c1[i], cnt - 1);} else {map.remove(c1[i]);}} else {return false;}}return true;}
}

解法2：由于本题只有小写字母，所以根据这个特性，可以用数组代替 hashmap

class Solution {public boolean canConstruct(String ransomNote, String magazine) {char[] chars1 = ransomNote.toCharArray();char[] chars2 = magazine.toCharArray();if (chars1.length > chars2.length) {return false;}int[] rans = new int[26];for (char c : chars1) {rans[c - 'a']++;}for (char c : chars2) {rans[c - 'a']--;}for (int i = 0; i < rans.length; i++) {if (rans[i] > 0) {return false;}}return true;}
}

387. 字符串中的第一个唯一字符

https://leetcode-cn.com/problems/first-unique-character-in-a-string/

class Solution {public int firstUniqChar(String s) {int[] chars = new int[26];for (int i = 0; i < s.length(); i++) {chars[s.charAt(i) - 'a']++;}for (int i = 0; i < s.length(); i++) {if (chars[s.charAt(i) - 'a'] == 1) {return i;}}return -1;}
}

评论区有人说，可以建立一个索引表记录首次出现的位置，就不用遍历两次字符串了。

389. 找不同

https://leetcode-cn.com/problems/find-the-difference/

按位异或

class Solution {public char findTheDifference(String s, String t) {char res = 0;char[] c1 = s.toCharArray();char[] c2 = t.toCharArray();for (char c : c1) {res = (char) (res ^ c);}for (char c : c2) {res = (char) (res ^ c);}return res;}
}

392. 判断子序列

https://leetcode-cn.com/problems/is-subsequence/
双指针，最后如果 p1 走到末尾，即可说明是子序列

class Solution {public boolean isSubsequence(String s, String t) {int p1 = 0, p2 = 0;char[] c1 = s.toCharArray();char[] c2 = t.toCharArray();while (p1 < c1.length && p2 < c2.length) {if (c1[p1] == c2[p2]) {p1++;p2++;} else {p2++;}}return p1 == c1.length;}
}

401. 二进制手表

https://leetcode-cn.com/problems/binary-watch/

逆向思考：遍历所有的时间，留下相加和为num的

class Solution {public List<String> readBinaryWatch(int num) {List<String> resList = new ArrayList<>();for (int i = 0; i < 12; ++i)for (int j = 0; j < 60; ++j)if (Integer.bitCount((i << 6) | j) == num)resList.add(i + ":" + (j > 9 ? "" : "0") + j);return resList;}
}

404. 左叶子之和

https://leetcode-cn.com/problems/sum-of-left-leaves/

二叉树问题，递归

class Solution {public int sumOfLeftLeaves(TreeNode root) {return sumLeft(root, false);}public int sumLeft(TreeNode node, boolean isLeft) {if (node == null) return 0;// 是叶子且在左边if (node.left == null && node.right == null && isLeft) {return node.val + sumLeft(node.right, false);}// 不是叶子或不在左边return sumLeft(node.left, true) + sumLeft(node.right, false);}
}

405. 数字转换为十六进制数

https://leetcode-cn.com/problems/convert-a-number-to-hexadecimal/

class Solution {public String toHex(int num) {int base = 16;// list 存储进制转换后的每一位List<Integer> list = new ArrayList<Integer>();long longNum;// 将负的 num 处理为补码对应的数if (num == 0) {return "0";} else if (num < 0) {longNum = 2L * Integer.MAX_VALUE + num + 2;} else {longNum = num;}// 经过上面的转化，此时正负可以统一处理while (longNum != 0) {int t = (int) (longNum % base);list.add(0, t);longNum /= base;}// 转换成字符串StringBuilder sb = new StringBuilder();for (int n : list) {if (n > 9) sb.append((char) (n + 'a' - 10));else sb.append(n);}return sb.toString();}
}

409. 最长回文串

别人那天并不是统计最长回文串，而是可以让字母重新组合！

思路：统计每个字母出现的次数，想象一个对称轴，偶数次可以放在对称轴两侧，奇数次只能作为对称轴，所以最终只能有一个出现奇数次数字母放在对称轴上。

class Solution {public int longestPalindrome(String s) {char[] cc = new char[122 - 65 + 1];char[] ss = s.toCharArray();for (char c : ss) {cc[c - 'A']++;}int total = 0;boolean hasSingle = false;for (char c : cc) {if (c % 2 == 1) hasSingle = true;total += c / 2;}total *= 2;if (hasSingle) total += 1;return total;}
}

412. Fizz Buzz

https://leetcode-cn.com/problems/fizz-buzz/

堪称最简单的一道题

class Solution {public List<String> fizzBuzz(int n) {List<String> list = new ArrayList();for(int i=1;i<=n;i++){if(i%3==0&&i%5==0)list.add("FizzBuzz");else if(i%3==0)list.add("Fizz");else if(i%5==0)list.add("Buzz");else list.add(""+i);}return list;}
}

414. 第三大的数

https://leetcode-cn.com/problems/third-maximum-number/

class Solution {public int thirdMax(int[] nums) {int flag1,flag2,flag3;long max1,max2,max3;flag1=flag2=flag3=0;max1=max2=max3=Long.MIN_VALUE;for(int i:nums){if((long)i>max1){max1=(long)i;flag1=1;}}for(int i:nums){if((long)i>max2&&(long)i<max1){max2=(long)i;flag2=1;}}for(int i:nums){if((long)i>max3&&(long)i<max2){max3=(long)i;flag3=1;}}if(flag1+flag2+flag3==3)return (int)max3;return (int)max1;}
}