字符串篇四

数字与字符间的转换

Integer.parseInt(a);将a转换成int类型

Integer.toString(a);将a转换成String类型

int[] cp = score.clone();将数组score的值赋给cp

299. 猜数字游戏 - 力扣（LeetCode）

• 这个题说白了还是统计字符类别的

class Solution {
    public String getHint(String secret, String guess) {
        int bulls = 0;
        int[] cntS = new int[10];
        int[] cntG = new int[10];
        for(int i = 0; i < secret.length(); i++){
            if(secret.charAt(i) == guess.charAt(i)) ++bulls;
            else{
                //这里相当于统计字符
                ++cntS[secret.charAt(i) - '0'];
                ++cntG[guess.charAt(i) - '0'];
            }
        }
        int cows = 0;
        for(int i = 0; i < 10; ++i){
            //理解一个点cntG[i], cntS[i]在同一位置上的字符肯定是相等的
            cows += Math.min(cntG[i], cntS[i]);
        }
        return Integer.toString(bulls) + "A" + Integer.toString(cows) + "B";
    }
}

412. Fizz Buzz - 力扣（LeetCode）

class Solution {
    public List<String> fizzBuzz(int n) {
        List<String> ans = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            String cur = "";
            if (i % 3 == 0) cur += "Fizz";
            if (i % 5 == 0) cur += "Buzz";
            if (cur.length() == 0) cur = i + "";
            ans.add(cur);
        }
        return ans;
    }
}

506. 相对名次 - 力扣（LeetCode）

class Solution {
    String[] ss = new String[]{"Gold Medal", "Silver Medal", "Bronze Medal"};
    public String[] findRelativeRanks(int[] score) {
        int n = score.length;
        String[] ret = new String[n];
        int[] clone = score.clone();
        Arrays.sort(clone);
        Map<Integer, Integer> map = new HashMap<>();
        for(int i = n - 1; i >= 0; i--) map.put(clone[i], n - 1 - i);
        for(int i = 0; i < n; i++){
            int rank = map.get(score[i]);
            ret[i] = rank < 3 ? ss[rank] : String.valueOf(rank + 1);
        }
        return ret;
    }
}

539. 最小时间差 - 力扣（LeetCode）

class Solution {
    public int findMinDifference(List<String> timePoints) {
        int n = timePoints.size() * 2;
        int[] nums = new int[n];
        for (int i = 0, idx = 0; i < n / 2; i++, idx += 2) {
            String[] ss = timePoints.get(i).split(":");
            int h = Integer.parseInt(ss[0]), m = Integer.parseInt(ss[1]);
            nums[idx] = h * 60 + m;
            nums[idx + 1] = nums[idx] + 1440;
        }
        Arrays.sort(nums);
        int ans = nums[1] - nums[0];
        for (int i = 0; i < n - 1; i++) ans = Math.min(ans, nums[i + 1] - nums[i]);
        return ans;
    }
}	

substring（）当有两个参数的时候左闭右开区间，当只有一个参数的时候从这个参数开始(闭区间)一直到字符结束位置。

String ss = "abcdefg";
String e= ss.substring(0,3);
String c = ss.substring(5);
System.out.println("两个参数：" +e);             \\两个参数：abc
System.out.println("一个参数：" + c);            \\一个参数：fg

class Solution {
    public int findMinDifference(List<String> timePoints) {
        int[] arr = new int[timePoints.size()];
        for(int i = 0; i < arr.length; i++){
            // System.out.println(timePoints.get(i).substring(0, 2));
            // System.out.println(timePoints.get(i).substring(3));
            arr[i] = Integer.parseInt(timePoints.get(i).substring(0, 2))*60 + Integer.parseInt(timePoints.get(i).substring(3));
        }
        // System.out.println(Arrays.toString(arr));
        Arrays.sort(arr);
        int min = Integer.MAX_VALUE;
        for(int i = 1; i < arr.length; i++) min = Math.min(min, arr[i] - arr[i - 1]);
        
        return Math.min(min, arr[0] + 1440 - arr[arr.length - 1]);

    }
}

553. 最优除法 - 力扣（LeetCode）

class Solution {
    public String optimalDivision(int[] nums) {
        int n = nums.length;
        StringBuilder bs = new StringBuilder();
        for(int i = 0; i < n; i++){
            bs.append(nums[i]);
            if(i + 1 < n) bs.append("/");
        }
        if(n > 2){
            bs.insert(bs.indexOf("/") + 1, "(");
            bs.append(")");
        }
        return bs.toString();
    }
}

537. 复数乘法 - 力扣（LeetCode）

.split()方法的解释

String str1 = "Welcome-To-Shanxi-XiAn";
String[] ss1 = str1.split("-");
//只是限定了分隔符号-所以返回了：[Welcome, To, Shanxi, XiAn]
System.out.println(Arrays.toString(ss1));

String[] sss1 = str1.split("-", 2);
//分隔符设置分割份数返回值，因为限定了两个所以返回：[Welcome, To-Shanxi-XiAn]
System.out.println(Arrays.toString(sss1));

String str2 = "www.baidu.com";
String[] ss2 = str2.split("\\.");
//转义字符的返回值：[www, baidu, com]，如果不进行转义就会返回[]。
System.out.println(Arrays.toString(ss2));

String str3 = "acount=? and uu =? or n=?";
String[] ss3 = str3.split("and|or");
//多个分隔符返回值：[acount=? ,  uu =? ,  n=?]
System.out.println(Arrays.toString(ss3));

String str4 = "1+1i";
//通过\\进行转义并通过|进行分隔返回值：[1, 1]
String[] ss4 = str4.split("\\+|i");
System.out.println(Arrays.toString(ss4));

class Solution {
    public String complexNumberMultiply(String num1, String num2) {
        String[] ss1 = num1.split("\\+|i"), ss2 = num2.split("\\+|i");
        int a = parse(ss1[0]), b = parse(ss1[1]);
        int c = parse(ss2[0]), d = parse(ss2[1]);
        int A = a * c - b * d, B = b * c + a * d;
        return A + "+" + B + "i";
    }
    int parse(String s) {
        return Integer.parseInt(s);
    }
}

443. 压缩字符串 - 力扣（LeetCode）

代码整的没太懂

class Solution {
    public int compress(char[] cs) {
        int n = cs.length;
        int i = 0, j = 0;
        while (i < n) {
            int idx = i;
            while (idx < n && cs[idx] == cs[i]) idx++;
            int cnt = idx - i;
            cs[j++] = cs[i];
            if (cnt > 1) {
                int start = j, end = start;
                while (cnt != 0) {
                    cs[end++] = (char)((cnt % 10) + '0');
                    cnt /= 10;
                }
                reverse(cs, start, end - 1);
                j = end;
            }
            i = idx;
        }
        return j;
    }
    void reverse(char[] cs, int start, int end) {
        while (start < end) {
            char t = cs[start];
            cs[start] = cs[end];
            cs[end] = t;
            start++; end--;
        }
    }
}

13. 罗马数字转整数 - 力扣（LeetCode）

class Solution {
    public int romanToInt(String s) {
        int[] gra1 = new int[] {1, 5, 10, 50,100,500,1000};
        char[] gra2 = new char[]{'I', 'V', 'X', 'L', 'C', 'D', 'M'};
        Map<Character, Integer> map = new HashMap<>();
        for(int i = 0; i < 7; i++) map.put(gra2[i], gra1[i]);
        int n = s.length();
        int max = Integer.MIN_VALUE;
        int ans = 0;
        for(int i = n - 1; i > -1; i--){
            int v = map.get(s.charAt(i));
            if(v >= max){
                ans += v;
                max = v;
            }else{
                ans -= v;
            }
        }
        return ans;
    }
}

12. 整数转罗马数字 - 力扣（LeetCode）

class Solution {
    public String intToRoman(int num) {
        int[] value = new int[] {1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900,1000};
        String[] key = new String[]{"I", "IV", "V", "IX", "X", "XL", "L", "XC", "C","CD","D", "CM","M"};
        StringBuilder ret = new StringBuilder();
        for(int i =value.length - 1; i > -1; i--){
            int val = value[i];
            String symbol = key[i];
            while(num >= val){
                num -= val;
                ret.append(symbol);
            }
            if(num == 0) break;
        }
        return ret.toString();
    }
}

481. 神奇字符串 - 力扣（LeetCode）

class Solution {
    public int magicalString(int n) {
        //这里定义n + 2大小的目的是防止while循环里j++加了两次而造成数组越界
        int[] cnt = new int[n + 2];
        cnt[0] = 1; cnt[1] = cnt[2] = 2;
        int j = 3, i = 2, c = 2;
        while(j < n){
            c ^= 3;
            cnt[j++] = c;
            if(cnt[i++] == 2) cnt[j++] = c;
        }
        int ans = 0;
        for(int b = 0; b < n; b++) ans += 2 - cnt[b];
        return ans;
    }
}

上面的这种方法太过于难以像到了，尤其是 C ^= 3 这一句。使用下面进行模拟就比较容易理解

class Solution {
	public int magicalString(int n) {
		StringBuilder bs = new StringBuilder("122");
		String BLX = "1";
		//这里注意一定是从2开始的，脑子又不灵光了
		int x = 2;
		while (x < n){
			if (bs.charAt(x) == '1') bs.append(BLX.repeat(1));
			else bs.append(BLX.repeat(2));
            //这一句这样写感觉有一点丑陋
			BLX = BLX == "1" ? "2" : "1";
            //BLX = BLX.equals("1") ? "2" : "1";
			x++;
		}
		System.out.println(bs.toString());
		int ans = 0;
		for (int i = 0; i < n; i++) ans += '2' - bs.charAt(i) ;
		return ans;
	}
}

上面的这种方法对于方法的使用和利用很好