lintcode做题总结, Oct 31 B

最新推荐文章于 2024-07-22 12:52:52 发布

原创最新推荐文章于 2024-07-22 12:52:52 发布 · 429 阅读

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本文解析了三道LeetCode经典题目：LargestNumber、MinimumWindowSubstring及ReverseLinkedList的解决方案，涉及数组排序、双指针技巧、哈希表、链表反转等核心算法。

1. Largest Number 这一题技巧性很强，对数组排序根据a+b和b+a进行比较。

public class Solution {
    public String largestNumber(int[] nums) {
        String[] strs = new String[nums.length];
        for(int i = 0; i < nums.length; i++){
            strs[i] = String.valueOf(nums[i]);
        }
     
        Arrays.sort(strs, new Comparator<String>(){
            public int compare(String s1, String s2){
                String leftRight = s1 + s2;
                String rightLeft = s2 + s1;
                return rightLeft.compareTo(leftRight);
     
            }
        });
     
        StringBuilder sb = new StringBuilder();
        for(String s : strs){
            sb.append(s);
        }
     
        while(sb.charAt(0) == '0' && sb.length() > 1){
            sb.deleteCharAt(0);
        }
     
        return sb.toString();
    }
}

2. Minimum Window Substring 使用2个map座位window

public class Solution {
    public String minWindow(String s, String t) {
        if(t.length()>s.length()) 
            return "";
        String result = "";
        //character counter for t
        HashMap<Character, Integer> target = new HashMap<Character, Integer>();
        for(int i = 0; i < t.length(); i++){
            char c = t.charAt(i);
            if (target.containsKey(c)) {
                target.put(c, target.get(c) + 1);
            } else {
                target.put(c, 1);  
            }
        }
        // character counter for s
        HashMap<Character, Integer> map = new HashMap<Character, Integer>();
        int left = 0;
        int minLen = s.length() + 1;
        int count = 0; // the total of mapped characters
        for(int i = 0; i < s.length(); i++){
            char c = s.charAt(i);
            if(target.containsKey(c)){
                if(map.containsKey(c)){
                    if(map.get(c) < target.get(c)){
                        count++;
                    }
                    map.put(c,map.get(c)+1);
                }else{
                    map.put(c,1);
                    count++;
                }
            }
            
            if(count == t.length()){
                char sc = s.charAt(left);
                while (!map.containsKey(sc) || map.get(sc) > target.get(sc)) {
                    if (map.containsKey(sc) && map.get(sc) > target.get(sc))
                        map.put(sc, map.get(sc) - 1);
                    left++;
                    sc = s.charAt(left);
                }
                if (i - left + 1 < minLen) {
                    result = s.substring(left, i + 1);
                    minLen = i - left + 1;
                }
            }
        }
        return result;
    }
}

3. Reverse Linked List 下面是两种不同的添加反转方法

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseList(ListNode head) {
        if (head == null) {
            return null;
        }
        /*
        ListNode start = new ListNode(0);
        start.next = head;
        ListNode p = head;
        ListNode q = p.next;
        while (p.next != null) {
            q = p.next.next;
            p.next.next = start.next;
            start.next = p.next;
            p.next = q;
        }
        return start.next;
        */
        /* for example : 432156
        * head:4, p:1, q:5, r:6
        */
        ListNode p = head, q, r;
        while (p.next != null) {
            q = p.next;
            r = q.next;
            q.next = head;
            p.next = r;
            head = q;
        }
        return head;
    }
}