算法通关村第十二关|黄金挑战|最长公共前缀&字符串压缩

1.最长公共前缀

原题：力扣14.

1.从前到后比较每个字符串的同一个位置。

public String longestCommonPrefix(String[] strs) {
	if (strs == null || strs.length == 0) {
        return "";
    }
    int length = strs[0].length();
    int count = strs.length;
    for (int i = 0; i < length; i++) {
        char c = strs[0].charAt(i);
        for (int j = 1; j < count; j++) {
            if (i == strs[j].length() || strs[j].charAt(i) != c) {
                return strs[0].substring(0, i);
            }
        }
    }
    return strs[0];
}

2.将完整的字符串进行比较得到公共前缀。然后遍历每个字符串，不断缩小这个公共前缀，遍历结束后得到的就是最长的公共前缀。

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++) {
        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++;
    }
    return str1.substring(0, index);
}

2.字符串压缩问题

原题：力扣443.

public int compress(char[] chars) {
	int n = chars.length;
    int write = 0, left = 0;
    for (int read = 0; read < n; read++) {
        if (read == n - 1 || chars[read] != chars[read + 1]) {
            chars[write++] = chars[read];
            int num = read - left + 1;
            if (num > 1) {
                int anchor = write;
                while (num > 0) {
                    chars[write++] = (char)(num % 10 + '0');
                    num /= 10;
                }
                reverse(chars, anchor, write - 1);
            }
            left = read + 1;
        }
    }
    return write;
}
public void reverse(char[] chars, int left, int right) {
	while (left < right) {
        char temp = chars[left];
        chars[left] = chars[right];
        chars[right] = temp;
        left++;
        right--;
    }
}

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