leetcode 解题

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罗马数字转整数

Description

public int romanToInt(String s) {
    HashMap<String, Integer> map = new HashMap<>();
    map.put("I", 1);
    map.put("V", 5);
    map.put("X", 10);
    map.put("L", 50);
    map.put("C", 100);
    map.put("D", 500);
    map.put("M", 1000);

    if (s.length() == 1)
        return map.get(String.valueOf(s.charAt(0)));
    int sum = 0;
    int temp = map.get(String.valueOf(s.charAt(0)));
    int next = 0;
    int pre = 0;
    for (int i = 0; i < s.length() - 1; i++) {
        next = map.get(String.valueOf(s.charAt(i + 1)));
        pre = map.get(String.valueOf(s.charAt(i)));
        if (next > pre) {
            temp = next - temp;
        } else if (next == pre) {
            temp = next + temp;
        } else {
            sum = sum + temp;
            temp = next;
        }
    }

    return sum + temp;
}

有效的括号

Description

public boolean isValid(String s) {
    if (s.equals("")) return true;
    if (s.isEmpty() || s.length() % 2 != 0) return false;
    Stack<Character> stack = new Stack<>();
    char[] chars = s.toCharArray();
    for (char c : chars) {
        if (c == '{' || c == '[' || c == '(') {
            stack.push(c);
        } else {
            if (stack.isEmpty()) return false;
            if (c == '}' && !stack.pop().equals('{')) return false;
            if (c == ']' && !stack.pop().equals('[')) return false;
            if (c == ')' && !stack.pop().equals('(')) return false;
        }

    }
    return stack.isEmpty();
}

合并两个有序链表

Description

递归处理

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */

public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
    if (l1 == null) return l2;

    if (l2 == null) return l1;

    ListNode listNode;

    if (l1.val < l2.val) {
        listNode = l1;
        listNode.next = mergeTwoLists(l1.next, l2);
    } else {
        listNode = l2;
        listNode.next = mergeTwoLists(l1, l2.next);
    }
    return listNode;
}