标签：pos keyset distinct row efi hashmap ini open []
kth-smallest-number-in-sorted-matrix
Find the kth smallest number in at row and column sorted matrix.
[
[1 ,5 ,7],
[3 ,7 ,8],
[4 ,8 ,9],
]
注意点：
Comparator<Pair>的实现
PriorityQueue poll add
1 import java.util.*; 2 3 class Pair{ 4 public int x, y, val; 5 public Pair(int x, int y, int val){ 6 this.x = x; 7 this.y = y; 8 this.val = val; 9 } 10 } 11 12 class PairComparator implements Comparator<Pair> { 13 public int compare(Pair a, Pair b){ 14 return a.val - b.val; 15 } 16 } 17 18 public class Solution { 19 /** 20 * @param matrix: a matrix of integers 21 * @param k: an integer 22 * @return: the kth smallest number in the matrix 23 */ 24 public int kthSmallest(int[][] matrix, int k) { 25 // write your code here 26 int[] dx = {0, 1}; 27 int[] dy = {1, 0}; 28 int m = matrix.length; 29 int n = matrix[0].length; 30 boolean[][] visited = new boolean[m][n]; 31 PriorityQueue<Pair> pq = new PriorityQueue<>(k, new PairComparator()); 32 pq.add(new Pair(0, 0, matrix[0][0])); 33 34 for (int i = 0; i < k - 1; i++){ 35 Pair cur = pq.poll(); 36 for (int j = 0; j < 2; j++){ 37 int x = cur.x + dx[j]; 38 int y = cur.y + dy[j]; 39 if (x < m && y < n && !visited[x][y]){ 40 pq.add(new Pair(x, y, matrix[x][y])); 41 visited[x][y] = true; 42 43 } 44 } 45 } 46 return pq.peek().val; 47 } 48 }
minimum-size-subarray-sum
Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn‘t one, return -1 instead.
注意点：
1 public class Solution { 2 /** 3 * @param nums: an array of integers 4 * @param s: an integer 5 * @return: an integer representing the minimum size of subarray 6 */ 7 public int minimumSize(int[] nums, int s) { 8 // write your code here 9 if (nums == null || nums.length == 0) return -1; 10 int result = -1; 11 int j = 0; 12 int sum = 0; 13 for (int i = 0; i < nums.length; i++){ 14 boolean flag = false; 15 if (i > 0){ 16 sum -= nums[i - 1]; 17 flag = true; 18 } 19 while (j < nums.length){ 20 if (!flag) { 21 sum += nums[j]; 22 } 23 else { 24 flag = false; 25 } 26 if (sum >= s){ 27 if (result == -1 || j - i + 1 < result){ 28 result = j - i + 1; 29 } 30 break; 31 } 32 j++; 33 } 34 } 35 return result; 36 } 37 }
longest-substring-without-repeating-characters
Given a string, find the length of the longest substring without repeating characters.
注意点：
1 import java.util.*; 2 public class Solution { 3 /** 4 * @param s: a string 5 * @return: an integer 6 */ 7 public int lengthOfLongestSubstring(String s) { 8 // write your code here 9 int result = 0; 10 if (s == null || s.length() == 0){ 11 return result; 12 } 13 Map<Character, Integer> map = new HashMap<>(); 14 int j = 0; 15 for (int i = 0; i < s.length(); i++){ 16 if (i > 0){ 17 map.put(s.charAt(i - 1), map.get(s.charAt(i - 1)) - 1); 18 if (map.get(s.charAt(i - 1)) == 0){ 19 map.remove(s.charAt(i - 1)); 20 } 21 } 22 while(j < s.length()){ 23 if (map.containsKey(s.charAt(j))){ 24 break; 25 } 26 else { 27 map.put(s.charAt(j), 1); 28 if (map.entrySet().size() > result){ 29 result = map.entrySet().size(); 30 } 31 j++; 32 } 33 } 34 } 35 return result; 36 } 37 }
longest-substring-with-at-most-k-distinct-characters
Given a string s, find the length of the longest substring T that contains at most k distinct characters.
注意：
map.entrySet().size() > k 判断键的个数也可以用keySet
1 public class Solution { 2 /** 3 * @param s : A string 4 * @return : The length of the longest substring 5 * that contains at most k distinct characters. 6 */ 7 public int lengthOfLongestSubstringKDistinct(String s, int k) { 8 // write your code here 9 int result = 0; 10 if (s == null || s.length() == 0){ 11 return result; 12 } 13 Map<Character, Integer> map = new HashMap<>();a 14 int j = 0; 15 for (int i = 0; i < s.length(); i++){ 16 boolean flag = false; 17 if (i > 0){ 18 map.put(s.charAt(i - 1), map.get(s.charAt(i - 1)) - 1); 19 if (map.get(s.charAt(i - 1)) == 0){ 20 map.remove(s.charAt(i - 1)); 21 } 22 flag = true; 23 } 24 while(j < s.length()){ 25 if(!flag){ 26 if (map.containsKey(s.charAt(j))){ 27 map.put(s.charAt(j), map.get(s.charAt(j)) + 1); 28 } 29 else{ 30 map.put(s.charAt(j), 1); 31 } 32 } 33 else { 34 flag = false; 35 } 36 if (map.entrySet().size() > k){ 37 break; 38 } 39 else { 40 if (j - i + 1 > result){ 41 result = j - i + 1; 42 } 43 j++; 44 } 45 } 46 } 47 return result; 48 } 49 }
kth-smallest-sum-in-two-sorted-arrays
Given two integer arrays sorted in ascending order and an integer k. Define sum = a + b, where a is an element from the first array and b is an element from the second one. Find the kth smallest sum out of all possible sums.
注意点：
1 import java.util.*; 2 3 class Pair{ 4 public int x, y, val; 5 public Pair(int x, int y, int val){ 6 this.x = x; 7 this.y = y; 8 this.val = val; 9 } 10 } 11 12 class PairComparator implements Comparator<Pair>{ 13 public int compare(Pair a, Pair b){ 14 return a.val - b.val; 15 } 16 } 17 18 19 public class Solution { 20 /** 21 * @param A an integer arrays sorted in ascending order 22 * @param B an integer arrays sorted in ascending order 23 * @param k an integer 24 * @return an integer 25 */ 26 public int kthSmallestSum(int[] A, int[] B, int k) { 27 // Write your code here 28 int[] dx = {0, 1}; 29 int[] dy = {1, 0}; 30 int m = A.length; 31 int n = B.length; 32 boolean[][] visited = new boolean[m][n]; 33 PriorityQueue<Pair> pq = new PriorityQueue<>(k, new PairComparator()); 34 pq.add(new Pair(0, 0, A[0] + B[0])); 35 for (int i = 0; i < k - 1; i++){ 36 Pair cur = pq.poll(); 37 for(int j = 0; j < 2; j++){ 38 int x = cur.x + dx[j]; 39 int y = cur.y + dy[j]; 40 if (x < m && y < n && !visited[x][y]){ 41 visited[x][y] = true; 42 pq.add(new Pair(x, y, A[x] + B[y])); 43 } 44 } 45 } 46 return pq.peek().val; 47 } 48 }
kth-largest-in-n-arrays
Find K-th largest element in N arrays.
N个数组找第K大
1 import java.util.*; 2 3 class Item{ 4 public int val, index, array; 5 public Item(int val, int index, int array){ 6 this.val = val; 7 this.index = index; 8 this.array = array; 9 } 10 } 11 12 class ItemComparator implements Comparator<Item>{ 13 public int compare(Item a, Item b){ 14 return b.val - a.val; 15 } 16 } 17 18 public class Solution { 19 /** 20 * @param arrays a list of array 21 * @param k an integer 22 * @return an integer, K-th largest element in N arrays 23 */ 24 public int KthInArrays(int[][] arrays, int k) { 25 // Write your code here 26 PriorityQueue<Item> pq = new PriorityQueue<>(k, new ItemComparator()); 27 for (int i = 0; i < arrays.length; i++){ 28 Arrays.sort(arrays[i]); 29 if (arrays[i].length > 0){ 30 pq.add(new Item(arrays[i][arrays[i].length - 1], arrays[i].length - 1, i)); 31 } 32 } 33 for (int i = 0; i < k - 1; i++){ 34 Item item = pq.poll(); 35 if (item.index > 0){ 36 pq.offer(new Item(arrays[item.array][item.index - 1], item.index - 1, item.array)); 37 } 38 } 39 return pq.peek().val; 40 } 41 }
two-sum-less-than-or-equal-to-target
Given an array of integers, find how many pairs in the array such that their sum is less than or equal to a specific target number. Please return the number of pairs.
找小于等于target的pairs数
注意点：
1 import java.util.*; 2 public class Solution { 3 /** 4 * @param nums an array of integer 5 * @param target an integer 6 * @return an integer 7 */ 8 public int twoSum5(int[] nums, int target) { 9 // Write your code here 10 int result = 0; 11 if (nums == null || nums.length < 2){ 12 return result; 13 } 14 Arrays.sort(nums); 15 int left = 0; 16 int right = nums.length - 1; 17 while (left < right){ 18 if (nums[left] + nums[right] <= target){ 19 result += right - left; 20 left++; 21 } 22 else{ 23 right--; 24 } 25 } 26 return result; 27 } 28 }
triangle-count
Given an array of integers, how many three numbers can be found in the array, so that we can build an triangle whose three edges length is the three numbers that we find?
上一题的变形题，先排序，固定最后一个元素，找前面大于target的pairs数
1 import java.util.*; 2 public class Solution { 3 /** 4 * @param S: A list of integers 5 * @return: An integer 6 */ 7 public int triangleCount(int S[]) { 8 // write your code here 9 Arrays.sort(S); 10 int result = 0; 11 for(int i = S.length - 1; i > 1; i--){ 12 result += twoSum(S, 0, i - 1, S[i]); 13 } 14 return result; 15 } 16 17 private int twoSum(int S[], int start, int end, int target){ 18 int result = 0; 19 while (start < end){ 20 if (S[start] + S[end] > target){ 21 result += end - start; 22 end--; 23 } 24 else{ 25 start++; 26 } 27 } 28 return result; 29 } 30 }
标签：pos keyset distinct row efi hashmap ini open []
原文地址：http://www.cnblogs.com/zcy-backend/p/6950730.html
