1. 两数之和

java 复制代码

// 先求差，再查哈希表
public int[] twoSum(int[] nums, int target) {
    Map<Integer,Integer> map = new HashMap<>();
    for(int i = 0;i<nums.length;i++){
        int key = target - nums[i];
        if(map.containsKey(key)){
            return new int[]{map.get(key),i};
        }
        map.put(nums[i],i);
    }
    return new int[0];
}

2. 两数相加

java 复制代码

	// 对应位置相加，记录进位，然后链表尾插法即可
	public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int flag = 0,lv1,lv2;
        ListNode answer = null,target = null;
        while (l1 != null || l2 != null){
            lv1 = l1 == null ? 0:l1.val;
            lv2 = l2 == null ? 0:l2.val;
            l1 = l1 == null ? null:l1.next;
            l2 = l2 == null ? null:l2.next;
            int sum = lv1+lv2+flag;
            flag = sum / 10;
            ListNode listNode = new ListNode(sum % 10);
            if (target == null){
                target = listNode;
                answer = target;
            }else {
                target.next = listNode;
                target = target.next;
            }
        }
        if (flag >0){
            target.next = new ListNode(flag);
        }
        return answer;
    }

3. 无重复字符的最长字串

java 复制代码

	// 滑动窗口
	public int lengthOfLongestSubstring(String s){
        Set<Character> set = new HashSet<>();
        int start = 0,end = 0,answer=0;
        while (end < s.length()){
            if (set.contains(s.charAt(end))){
                set.remove(s.charAt(start++));
            }else {
                set.add(s.charAt(end++));
                answer = Math.max(answer,end - start);
            }
        }
        return answer;
    }

4. 最长回文子串

java 复制代码

 // 动态规划
 public String longestPalindrome(String s) {
        if (s == null || s.length() < 2) {
            return s;
        }
        int strLen = s.length();
        int maxStart = 0;  //最长回文串的起点
        int maxEnd = 0;    //最长回文串的终点
        int maxLen = 1;  //最长回文串的长度

        boolean[][] dp = new boolean[strLen][strLen];

        for (int r = 1; r < strLen; r++) {
            for (int l = 0; l < r; l++) {
                if (s.charAt(l) == s.charAt(r) && (r - l <= 2 || dp[l + 1][r - 1])) {
                    dp[l][r] = true;
                    if (r - l + 1 > maxLen) {
                        maxLen = r - l + 1;
                        maxStart = l;
                        maxEnd = r;
                    }
                }
            }
        }
        return s.substring(maxStart, maxEnd + 1);
    }