【代码随想录训练营】Day8-字符串

代码随想录 Day8

今日任务

344.反转字符串
541.反转字符串Ⅱ
剑指Offer 05.替换空格
151.反转字符串中的单词
剑指Offer58-Ⅱ.左旋转字符串
语言：Java

344. 反转字符串

考点：字符串
链接：https://leetcode.cn/problems/reverse-string/

class Solution {
    public void reverseString(char[] s) {
        int left = 0;
        int right = s.length - 1;
        while(left < right){
            char temp = s[left];
            s[left] = s[right];
            s[right] = temp;
            left++;
            right--;
        }
    }
}

541. 反转字符串Ⅱ

考点：字符串
链接：https://leetcode.cn/problems/reverse-string-ii/
注意事项：字符串常量要先转为字符数组再进行后续操作

class Solution {
    public void myReverse(char[] cArray, int index1, int index2){
        int left = index1;
        int right = index2;
        while(left < right){
            char temp = cArray[left];
            cArray[left] = cArray[right];
            cArray[right] = temp;
            left++;
            right--;
        }
    }

    public String reverseStr(String s, int k) {
        char[] cArray = s.toCharArray();
        for(int i = 0; i < s.length();){
            int left = s.length() - i;
            if(left < k){
                myReverse(cArray, i, s.length() - 1);
                i += left;
            }
            else if(left >= k && left < 2 * k){
                myReverse(cArray, i, i + k - 1);
                i += left;
            }
            else if(left >= 2 * k){
                myReverse(cArray, i, i + k - 1);
                i += 2 * k;
            }
        }
        return new String(cArray);
    }
}

剑指Offer 05.替换空格

考点：字符串
链接：https://leetcode.cn/problems/ti-huan-kong-ge-lcof/
改进：
① 直接用 StringBuffer / StringBuilder 遍历
② 双指针，先用 StringBuffer / StringBuilder 遍历出空格数量；然后left 指向原字符串尾，right 指向填充空格后的字符串尾；将 s 转为字符数组，left 和 right 同时向左移动

class Solution {
    public String replaceSpace(String s) {
        // 会产生过多冗余空间
        char[] cArray = new char[s.length() * 3];
        int j = 0;
        for(int i = 0; i < s.length(); i++){
            if(s.charAt(i) != ' '){
                cArray[j++] = s.charAt(i);
            }
            else if(s.charAt(i) == ' '){
                cArray[j++] = '%';
                cArray[j++] = '2';
                cArray[j++] = '0';
            }
        }
        return new String(cArray, 0, j);
    }
}

151. 反转字符串中的单词

考点：字符串
链接：https://leetcode.cn/problems/reverse-words-in-a-string/
改进：先去除多余空格，反转整个字符串，反转每个单词

class Solution {
    public String reverseWords(String s) {
        StringBuffer sb = new StringBuffer();
        int left, right;
        for(int i = s.length() - 1; i >= 0;){
            while(i >= 0 && s.charAt(i) == ' '){
                i--;
            }
            right = i;
            while(i >= 0 && s.charAt(i) != ' '){
                i--;
            }
            //这里的if判断不可以写成 if(i >= 0)
            //否则对于以字母非空格开头的字符串会忽略掉起始单词
            //if(i >= 0){
            if(right >= 0){
                left = i + 1;
                sb.append(s.substring(left, right + 1));
                sb.append(' ');
                right = i;
            }
        }
        sb.delete(sb.length() - 1, sb.length());
        return sb.toString();
    }
}

剑指Offer 58-Ⅱ.左旋转字符串

考点：字符串
链接：https://leetcode.cn/problems/zuo-xuan-zhuan-zi-fu-chuan-lcof/
直接使用自带substring函数是没有意义的

class Solution {
    public String reverseLeftWords(String s, int n) {
        if(n >= s.length()){
            return s;
        }
        StringBuffer sb = new StringBuffer();
        sb.append(s.substring(n, s.length()));
        sb.append(s.substring(0, n));
        return sb.toString();
    }
}

改进：将前n个字符转移到字符串尾部就相当于在原字符串中先反转前n个元素，再反转其余元素，最后将整个字符串反转

class Solution {
    public void reverse(char[] cArray, int index1, int index2){
        int left = index1;
        int right = index2;
        while(left < right){
            char temp = cArray[left];
            cArray[left] = cArray[right];
            cArray[right] = temp;
            left++;
            right--;
        }
    }
    public String reverseLeftWords(String s, int n) {
        char[] cArray = s.toCharArray();
        reverse(cArray, 0, n - 1);
        reverse(cArray, n, s.length() - 1);
        reverse(cArray, 0, s.length() - 1);
        return new String(cArray);
    }
}