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最新推荐文章于 2022-11-25 16:10:47 发布

转载最新推荐文章于 2022-11-25 16:10:47 发布 · 66 阅读

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/lizzyluvcoding/p/10772136.html

本文深入解析双指针技术在解决最小子数组、最长无重复子串等问题的应用，以及堆在处理多数组中第K大元素、矩阵中第K小元素等场景的高效算法实现。

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406. Minimum Size Subarray

同向双指针

https://www.lintcode.com/problem/minimum-size-subarray-sum/description?_from=ladder&&fromId=4

 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){
10             return -1;
11         }
12         
13         int sum = 0;
14         int res = Integer.MAX_VALUE;
15         for(int l =0,r=0;r<nums.length;r++){
16             sum +=nums[r];
17             
18             while(sum>=s){
19                 res = Math.min(res,r-l+1);
20                 sum = sum -nums[l];
21                 l++;
22             }
23         }
24         
25         return res==Integer.MAX_VALUE?-1:res;
26     }
27 }

View Code

384. Longest Substring Without Repeating Characters

同向双指针

https://www.lintcode.com/problem/longest-substring-without-repeating-characters/description?_from=ladder&&fromId=4

 1 public class Solution {
 2     /**
 3      * @param s: a string
 4      * @return: an integer
 5      */
 6     public int lengthOfLongestSubstring(String s) {
 7         // write your code here
 8         if(s==null || s.length()==0){
 9             return 0;
10         }
11         
12         Set<Character> set = new HashSet<>();
13         int left =0;
14         int right =0;
15         int ans = Integer.MIN_VALUE;
16         
17         while(left<s.length() && right<s.length()){
18             while(right<s.length() && !set.contains(s.charAt(right))){
19                 set.add(s.charAt(right));
20                 ans = Math.max(ans,right-left+1);
21                 right++;
22             }
23             
24             set.remove(s.charAt(left));
25             left++;
26         }
27         
28         return ans == Integer.MIN_VALUE ? -1 : ans;
29     }
30 }

View Code

32. Minimum Window Substring

同向双指针

https://www.lintcode.com/problem/minimum-window-substring/description

 1 public class Solution {
 2     /**
 3      * @param source : A string
 4      * @param target: A string
 5      * @return: A string denote the minimum window, return "" if there is no such a string
 6      */
 7     public String minWindow(String source , String target) {
 8         // write your code here    
 9         if(source==null || source.length()==0){
10             return source;
11         }
12         
13         Map<Character,Integer> map = new HashMap<>();
14         for(int i=0;i<target.length();i++){
15             char c = target.charAt(i);
16             if(map.containsKey(c)){
17                 map.put(c,map.get(c)+1);
18             }else{
19                 map.put(c,1);
20             }
21         }
22         
23         int i=0;
24         int j =0;
25         String res= "";
26         int min = Integer.MAX_VALUE;
27         int countT = target.length();
28         int countS = 0;
29         
30         //while 循环不要加 j<source.length 的条件
31         while(i<source.length()){
32             while(j<source.length() && countS<countT){
33                 char c = source.charAt(j);
34                 if(map.containsKey(c)){
35                     if(map.get(c)>0) countS++;
36                     map.put(c,map.get(c)-1);
37                 }
38                 
39                 j++;
40             }
41             
42             if(countS>=countT){
43                 if(j-i<min){
44                     min = j-i;
45                     res = source.substring(i,j);
46                 }
47             }
48             
49             char cc = source.charAt(i);
50             if(map.containsKey(cc)){
51                 if(map.get(cc)>=0) countS--;
52                 map.put(cc,map.get(cc)+1);
53             }
54             i++;
55         }
56         
57         return res;
58     }
59 }

View Code

386. Longest Substring with At Most K Distinct Characters

同向双指针

https://www.lintcode.com/problem/longest-substring-with-at-most-k-distinct-characters/description?_from=ladder&&fromId=4

 1 public class Solution {
 2     /**
 3      * @param s: A string
 4      * @param k: An integer
 5      * @return: An integer
 6      */
 7     public int lengthOfLongestSubstringKDistinct(String s, int k) {
 8         // write your code here
 9         if(s==null || s.length()==0 || k==0){
10             return 0;
11         }
12         
13         int l = 0;
14         int r = 0;
15         int ans = 0;
16         int sum = 0;
17         
18         int[] cnt = new int[256];
19         for(r=0;r<s.length();r++){
20             cnt[s.charAt(r)]++;
21             if(cnt[s.charAt(r)]==1){
22                 sum++;
23             }
24             
25             while(sum>k){
26                 cnt[s.charAt(l)]--;
27                 if(cnt[s.charAt(l)]==0){
28                     sum--;
29                 }
30                 l++;
31             }
32             
33             ans = Math.max(ans,r-l+1);
34         }
35         
36         return ans;
37         
38     }
39 }

View Code

465. Kth Smallest Sum In Two Sorted Arrays

heap

https://www.lintcode.com/problem/kth-smallest-sum-in-two-sorted-arrays/description?_from=ladder&&fromId=4

class Pair{
    int x;
    int y;
    int sum;
    Pair(int x, int y,int sum){
        this.x = x;
        this.y = y;
        this.sum = sum;
    }
}
public class Solution {
    /**
     * @param A: an integer arrays sorted in ascending order
     * @param B: an integer arrays sorted in ascending order
     * @param k: An integer
     * @return: An integer
     */
    public int kthSmallestSum(int[] A, int[] B, int k) {
        // write your code here
         if (A == null || A.length == 0) {
             return 0;
         }
         if (B == null || B.length == 0) {
             return 0;
         }
         if (k == 0) {
           return 0;
         }
         
         Comparator<Pair> comparator = new Comparator<Pair>(){
             @Override
             public int compare(Pair o1,Pair o2){
                 return o1.sum-o2.sum;
             }
         };
         PriorityQueue<Pair> minHeap = new PriorityQueue<Pair>(comparator);
         boolean[][] visit = new boolean[A.length][B.length];
         minHeap.add(new Pair(0,0,A[0]+B[0]));
         visit[0][0] = true;
         int[] dx = {0,1};
         int[] dy = {1,0};
         
         for(int i=1;i<k;i++){
             Pair center = minHeap.poll();;
             for(int j = 0;j<2;j++){
                 if(center.x + dx[j]>A.length-1 || center.y + dy[j]>B.length-1 || visit[center.x+dx[j]][center.y+dy[j]]){
                     continue;
                 }
                 minHeap.add(new Pair(center.x+dx[j],center.y+dy[j],A[center.x+dx[j]]+B[center.y+dy[j]]));
                 visit[center.x+dx[j]][center.y+dy[j]] = true;
             }
         }
         
         return minHeap.peek().sum;
    }
}

View Code

401. Kth Smallest Number in Sorted Matrix

heap

https://www.lintcode.com/problem/kth-smallest-number-in-sorted-matrix/description?_from=ladder&&fromId=4

 1 class Pair {
 2     private int x;
 3     private int y;
 4     private int val;
 5 
 6     public Pair(int x, int y, int val) {
 7         this.x = x;
 8         this.y = y;
 9         this.val = val;
10     }
11 
12     public int getX() {
13         return x;
14     }
15 
16     public void setX(int x) {
17         this.x = x;
18     }
19 
20     public int getY() {
21         return y;
22     }
23 
24     public void setY(int y) {
25         this.y = y;
26     }
27 
28     public int getVal() {
29         return val;
30     }
31 
32     public void setVal(int val) {
33         this.val = val;
34     }
35 }
36 
37 public class Solution {
38     /**
39      * @param matrix: a matrix of integers
40      * @param k:      An integer
41      * @return: the kth smallest number in the matrix
42      */
43     public int kthSmallest(int[][] matrix, int k) {
44         // write your code here
45         if (matrix == null || matrix.length == 0 || matrix.length * matrix[0].length < k) {
46             return -1;
47         }
48         Comparator<Pair> comparator = new Comparator<Pair>() {
49             @Override
50             public int compare(Pair o1, Pair o2) {
51                 return o1.getVal() - o2.getVal();
52             }
53         };
54         int r = matrix.length;
55         int c = matrix[0].length;
56         PriorityQueue<Pair> minHeap = new PriorityQueue<>(comparator);
57         boolean[][] visited = new boolean[r][c];
58 
59         minHeap.add(new Pair(0, 0, matrix[0][0]));
60         visited[0][0] = true;
61 
62         for (int i = 1; i <= k - 1; i++) {
63             Pair cur = minHeap.poll();
64             if (cur.getX() + 1 < r && !visited[cur.getX() + 1][cur.getY()]) {
65                 minHeap.add(new Pair(cur.getX() + 1, cur.getY(), matrix[cur.getX() + 1][cur.getY()]));
66                 visited[cur.getX() + 1][cur.getY()] = true;
67             }
68             if (cur.getY() + 1 < c && !visited[cur.getX()][cur.getY() + 1]) {
69                 minHeap.add(new Pair(cur.getX(), cur.getY() + 1, matrix[cur.getX()][cur.getY() + 1]));
70                 visited[cur.getX()][cur.getY() + 1] = true;
71             }
72         }
73 
74         return minHeap.peek().getVal();
75     }
76 }

View Code

543. Kth Largest in N Arrays

heap

https://www.lintcode.com/problem/kth-largest-in-n-arrays/description?_from=ladder&&fromId=4

 1 class Node{
 2     int value;
 3     int fromId;
 4     int index;
 5     public Node(int value,int fromId,int index){
 6         this.value = value;
 7         this.fromId = fromId;
 8         this.index = index;
 9     }
10 }
11 public class Solution {
12     /**
13      * @param arrays: a list of array
14      * @param k: An integer
15      * @return: an integer, K-th largest element in N arrays
16      */
17     public int KthInArrays(int[][] arrays, int k) {
18         // write your code here
19         if(arrays==null || arrays.length==0 ||k<=0){
20             return 0;
21         }
22         
23         Comparator<Node> comparator = new Comparator<Node>(){
24             @Override
25             public int compare(Node o1,Node o2){
26                 return o2.value - o1.value;
27             }
28         };
29         
30         PriorityQueue<Node> maxHeap = new PriorityQueue<Node>(comparator);
31         
32         int n = arrays.length;
33         
34         //sort and put first column into heap
35         for(int i =0;i<n;i++){
36             Arrays.sort(arrays[i]);
37             
38             if(arrays[i].length>0){
39                 int fromId = i;
40                 int index = arrays[i].length-1;
41                 int value = arrays[i][index];
42                 maxHeap.add(new Node(value,fromId,index));
43             }
44         }
45         
46         for(int i=1;i<k;i++){
47             Node curr = maxHeap.poll();
48             
49             if(curr.index>0){
50                 curr.index--;
51                 maxHeap.add(new Node(arrays[curr.fromId][curr.index],curr.fromId,curr.index));
52             }
53         }
54         
55         return maxHeap.peek().value;
56         
57         
58     }
59 }

View Code

转载于:https://www.cnblogs.com/lizzyluvcoding/p/10772136.html