OA1已过，七道debug非常简单都是原题，OA2 coding 70min两道题。准备如下
- Longest Palindromic Substring
Leetcode解法4，采用中心扩展法。
java:
public static String longestPalindrome(String s){
if(s == null || s.length() < 1) return "";
int start = 0, end = 0;
for(int i = 0; i < s.length(); i++){
int len1 = expandAroundCenter(s, i, i);
int len2 = expandAroundCenter(s, i, i+1);
int len = Math.max(len1, len2);
if(len > end - start + 1){
start = i - (len - 1) / 2;
end = i + len /2;
}
}
return s.substring(start, end + 1);
}
private static int expandAroundCenter(String s, int start, int end){
int L = start, R = end;
while(L >= 0 && R < s.length() && s.charAt(L) == s.charAt(R)){
L--;
R++;
}
return R-L-1;
}
执行结果： 通过 显示详情 执行用时 :8 ms, 在所有 java 提交中击败了95.86% 的用户 内存消耗 :36.5 MB, 在所有 java 提交中击败了90.70%的用户
- K-Closest Points to Origin
采用了比较弱的方法，时间复杂度O(NlogN)
java:
import java.util.Arrays;
public static int[][] kClosest(int[][] points, int K){
int N = points.length;
int[] distance = new int[N];
for(int i = 0; i < N; i++){
distance[i] = dist(points[i]);
}
Arrays.sort(distance);
int distofK = distance[K-1];
int[][] ans = new int[K][2];
int t = 0;
for(int i = 0; i < N; i++){
if(dist(points[i]) <= distofK){
ans[t++] = points[i];
}
}
return ans;
}
public static int dist(int[] p){
return p[0] * p[0] + p[1] * p[1];
}
执行结果： 通过 显示详情 执行用时 :14 ms, 在所有 java 提交中击败了87.90% 的用户 内存消耗 :62.8 MB, 在所有 java 提交中击败了84.80%的用户
- Path With Maximum Minimum Value
Given a matrix of integers A with R rows and C columns, find the maximum score of a path starting at [0,0] and ending at [R-1,C-1].
The score of a path is the minimum value in that path. For example, the value of the path 8 → 4 → 5 → 9 is 4.
A path moves some number of times from one visited cell to any neighbouring unvisited cell in one of the 4 cardinal directions (north, east, west, south).
Example 1:
Input: [[5,4,5],[1,2,6],[7,4,6]] Output: 4 Explanation: The path with the maximum score is highlighted in yellow.
Example 2:
Input: [[2,2,1,2,2,2],[1,2,2,2,1,2]] Output: 2
DFS:
java:
private row, col, max, min;
public static int maxMinPath(int[][] matrix){
row = matrix.length;
col = matrix[0].length;
max = Integer.MIN_VALUE;
min = Integer.MAX_VALUE;
dfshelper(matrix, min, 0, 0);
return max;
}
public static void dfshelper(int[][] matrix, int min, int i, int j){
if(i >= row || j >= col) return;
if(i == row - 1 && j == col - 1){
min = Math.min(min, matrix[i][j]);
max = Math.max(min, max);
}
min = Math.min(min, matrix[i][j]);
dfshelper(matrix, min, i, j+1);
dfshelper(matrix, min, i+1, j);
}
a) Two Sum
java:
public static int[] twoSum(int[] nums, int target){
HashMap<Integer, Integer> m = new HashMap<>();
for(int i = 0; i < nums.length; i++){
int diff = target - nums[i];
if(m.containsKey(diff))
return new int[]{m.get(diff), i};
m.put(nums[i], i);
}
throw new IllegalArgumentException();
}
执行结果： 通过 显示详情 执行用时 :3 ms, 在所有 java 提交中击败了98.87% 的用户 内存消耗 :37.3 MB, 在所有 java 提交中击败了90.54%的用户
b)
- Merge Two Sorted Linked Lists
java:
public ListNode mergeTwoLists(ListNode l1, ListNode l2){
if(l1 == null || l2 == null) return l1 == null ? l2 : l1;
ListNode dummy = new ListNode(0);
ListNode cur = dummy;
while(l1 != null || l2 != null){
int v1 = l1 == null ? Integer.MAX_VALUE : l1.val;
int v2 = l2 == null ? Integer.MAX_VALUE : l2.val;
if(v1 <= v2){
cur.next = l1;
cur = cur.next;
l1 = l1.next;
}
else{
cur.next = l2;
cur = cur.next;
l2 = l2.next;
}
}
return dummy.next;
}
执行结果： 通过 显示详情 执行用时 :1 ms, 在所有 java 提交中击败了92.63% 的用户 内存消耗 :39.4 MB, 在所有 java 提交中击败了68.61%的用户
- Most Common Word
java:
import java.util.*;
public class mostCommonWord { public static String mostCommonWord(String paragraph, String[] banned){ for(int i = 0; i< banned.length; i++){ banned[i] = banned[i].toLowerCase(); } for(int i = 0; i< banned.length; i++){ System.out.println(banned[i]); } Set set = new HashSet<>(Arrays.asList(banned)); set.add(""); paragraph += "."; String res = " "; int times = 0; Map<String, Integer> map = new HashMap<>(); int i = 0; for(int k = 0; k<paragraph.length(); k++){ char c = paragraph.charAt(k); if((c>='a' && c<='z') || (c>='A' && c <= 'Z')){ continue; } else{ String word = paragraph.substring(i, k); word = word.toLowerCase(); if(!set.contains(word)){ map.put(word, map.getOrDefault(word, 0)+1); if(map.get(word) > times){ res = word; times = map.get(word); } } i =k +1; }
}
return res;
}
public static void main(String[] args){
String paragraph = "Bob hit a ball, the hit BALL flew far after it was hit.";
String[] banned = {"hIJt"};
System.out.println(mostCommonWord(paragraph, banned));
}
}
- Substring of size k with k distinct characters
Given a string s and an int k, return all unique substrings of s of size k with k distinct characters.
Example 1:
Input: s = "abcabc", k = 3 Output: ["abc", "bca", "cab"]
Example 2:
Input: s = "abacab", k = 3 Output: ["bac", "cab"]
Example 3:
Input: s = "awaglknagawunagwkwagl", k = 4 Output: ["wagl", "aglk", "glkn", "lkna", "knag", "gawu", "awun", "wuna", "unag", "nagw", "agwk", "kwag"] Explanation: Substrings in order are: "wagl", "aglk", "glkn", "lkna", "knag", "gawu", "awun", "wuna", "unag", "nagw", "agwk", "kwag", "wagl" "wagl" is repeated twice, but is included in the output once.
Constraints:
The input string consists of only lowercase English letters [a-z]
0 ≤ k ≤ 26
