Description

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

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Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, two is written as II in Roman numeral, just two one’s added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

• I can be placed before V (5) and X (10) to make 4 and 9.
• X can be placed before L (50) and C (100) to make 40 and 90.
• C can be placed before D (500) and M (1000) to make 400 and 900.

Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

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Input: 3
Output: "III"

Example 2:

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Input: 4
Output: "IV"

Example 3:

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Input: 9
Output: "IX"

Example 4:

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Input: 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.

Example 5:

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Input: 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

Difficulty: Medium

Code:

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class Solution {
    public String intToRoman(int num) {
        
    }
}

题意

给定一个整数，将其转成罗马数字，整数的范围在1到3999。

Description

Given n non-negative integers a1, a2, …, an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

Example:

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Input: [1,8,6,2,5,4,8,3,7]
Output: 49

Difficulty: Medium

Code:

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class Solution {
    public int maxArea(int[] height) {
        
    }
}

题意

给定正整数1, a2, …, an，代表坐标轴横轴上1到n到地方，各有一条垂直到线段，其另一端到高度为正整数到值。
两根线及横轴组成了容器，求使容器装最多水到两条线段，并返回装水到面积。

Description

Given an input string (s) and a pattern (p), implement regular expression matching with support for ‘.’ and ‘*’.

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'.' Matches any single character.
'*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

Note:

• s could be empty and contains only lowercase letters a-z.
• p could be empty and contains only lowercase letters a-z, and characters like . or *.

Example 1:

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Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

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Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

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Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

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Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

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Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

Difficulty: Hard

Code:

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class Solution {
    public boolean isMatch(String s, String p) {
        
    }
}

题意

这道题关于正则表达式，’.’表示任意字符，’*’表示其之前的字符可以出现0次或者多次。要求返回字符串是否完全匹配正则表达式。

Description

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

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Input: 121
Output: true

Example 2:

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Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

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Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Difficulty: Easy

Code:

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class Solution {
    public boolean isPalindrome(int x) {

    }
}

题意

判断一个int数是否是回文格式，即从前读取和从后开始读取都是一样的。

Description

Implement atoi which converts a string to an integer.

The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned.

Note:

• Only the space character ‘ ‘ is considered as whitespace character.

• Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. If the numerical value is out of the range of representable values, INT_MAX (231 − 1) or INT_MIN (−231) is returned.

Example 1:

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Input: "42"
Output: 42

Example 2:

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Input: "   -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
             Then take as many numerical digits as possible, which gets 42.

Example 3:

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Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.

Example 4:

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Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical 
             digit or a +/- sign. Therefore no valid conversion could be performed.

Example 5:

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Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
             Thefore INT_MIN (−231) is returned.

Difficulty: Medium

Code:

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class Solution {
    public int myAtoi(String str) {
        
    }
}

题意

实现atoi函数将字符串转换为整数。

函数首先尽可能多的丢弃空白字符，直到发现第一个非空字符位为止。 接着从这个字符开始，读入一个可选的正负号，然后尽可能多的读入数字，最后将它们解析成数值。

字符串中在合法数字后可以包含额外的非法字符，对于这些字符只需丢弃即可。

如果字符串的非空字符不是一个有效的整数，或者，当字符串为空或者只包含空白字符时，不需要执行转换。

如果不能够执行有效的转换则返回0。如果得到的数值超出了整数范围，返回INT_MAX (2147483647) 或者 INT_MIN (-2147483648)。