字符串有三种编辑操作:插入一个字符、删除一个字符或者替换一个字符。 给定两个字符串，编写一个函数判定它们是否只需要一次(或者零次)编辑。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/one-away-lcci
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class Solution {

    private boolean judgeReplace(String first, String second) {
        int num = 0;
        for (int i = 0; i < first.length(); ++i) {
            if (first.charAt(i) != second.charAt(i)) {
                num++;
                if (num > 1) {
                    return false;
                }
            }
        }
        return true;
    }

    private boolean judgeInsert(String first, String second) {
        int p1 = 0, p2 = 0;
        boolean inserted = false;
        while (p2 < second.length()) {
            if (p1 == first.length() || first.charAt(p1) != second.charAt(p2)) {
                if (inserted) {
                    return false;
                }
                inserted = true;
                p2++;
            } else {
                p1++;
                p2++;
            }
        }
        return true;
    }

    private boolean judgeDelete(String first, String second) {
        return judgeInsert(second, first);
    }

    public boolean oneEditAway(String first, String second) {
        if (first.length() == second.length()) {
            return judgeReplace(first, second);
        } else if (first.length() < second.length()) {
            return judgeInsert(first, second);
        } else {
            return judgeDelete(first, second);
        }
    }
}