LeetCode-Java(03)

9. 回文数

class Solution {
    public boolean isPalindrome(int x) {
        if (x < 0 || (x % 10 == 0 && x != 0)) {
            return false;
        }

        int revertedNumber = 0;
        while (x > revertedNumber) {
            revertedNumber = revertedNumber * 10 + x % 10;
            x /= 10;
        }
        // 当长度为奇数时通过revertedNumber/10 去除处于中位的数字。
        return x == revertedNumber || x == revertedNumber / 10;
    }
}

11. 盛最多水的容器

双指针

class Solution {
    public int maxArea(int[] height) {
        int ans=0;
        // 首位指针
        int i=0,j=height.length-1;
        while(i=height[j]) j--;
            else i++;
        }
        return ans;
    }
}

写到一起 时间快不少

class Solution {
    public int maxArea(int[] height) {
        int ans=0;
        int i=0,j=height.length-1;
        while(i=height[j]) 
            ans=Math.max(ans,(j-i)*height[j--]);
            else 
            ans=Math.max(ans,(j-i)*height[i++]);
        }
        return ans;
    }
}

12. Integer to Roman

class Solution {
    public String intToRoman(int num) {
        // 4、9、40、90、400、900 作为加法因子，它们在结果中只能出现一次。
        int[] nums = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
        String[] romans = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};

        StringBuilder stringBuilder = new StringBuilder();
        int index = 0;
        while (index < 13) {
            // 贪心策略
            while (num >= nums[index]) {
                stringBuilder.append(romans[index]);
                num -= nums[index];
            }
            index++;
        }

        return stringBuilder.toString();
    }
}

13. Roman to Integer

和前一道相反，直接累加字符的值

class Solution {
    public int romanToInt(String s) {
        // 4、9、40、90、400、900 作为加法因子，它们在结果中只能出现一次。
        // int[] nums = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
        // String[] romans = {"M", "f", "D", "e", "C", "d", "L", "c", "X", "b", "V", "a", "I"};
        s = s.replace("IV","a");
        s = s.replace("IX","b");
        s = s.replace("XL","c");
        s = s.replace("XC","d");
        s = s.replace("CD","e");
        s = s.replace("CM","f");
        int res = 0;
        for (int i=0; i

优化一下

class Solution {
    public int romanToInt(String s) {
        int[] nums = new int[s.length()];
        int res = 0;
        for(int i = 0;i 

14. Longest Common Prefix

 每两个字符串进行比较

class Solution {
    public String longestCommonPrefix(String[] strs) {
        if (strs == null || strs.length == 0) {
            return "";
        }
        String prefix = strs[0];
        int count = strs.length;
        for (int i = 1; i < count; i++) {
            // 找两个字符串的公共最长前缀，两两比较，每次比较前面得出的最长前缀
            prefix = longestCommonPrefix(prefix, strs[i]);
            if (prefix.length() == 0) {
                break;
            }
        }
        return prefix;
    }

    public String longestCommonPrefix(String str1, String str2) {
        int length = Math.min(str1.length(), str2.length());
        int index = 0;
        while (index < length && str1.charAt(index) == str2.charAt(index)) {
            index++;
        }
        // substring 的参数索引 左闭右开
        return str1.substring(0, index);
    }
}