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package leetcode;

public class interview0105 {

/*

* first -->second

* 情况分三种：

* 1.替换 first.length()==second.length()

* 2.删除 first.length()=second.length()+1

* 3.添加 first.length()+1=second.length()

* */

public boolean oneEditAway(String first, String second) {

if (first.equals(second)) return true;

int len1 = Math.max(first.length(), second.length());

int len2 = Math.min(first.length(), second.length());

if (len1 - len2 > 1) return false;

if (len1 != first.length()) { //交换字符串，使得长字符串为 first

String tmp = first;

first = second;

second = tmp;

}

int count = 0;

if (len1 == len2) {

for (int i = 0; i < len1; i++) {

if (first.charAt(i) != second.charAt(i))

count++;

if (count > 1) return false;

}

} else {

//长短字符串 只允许有一处不匹配 例如 pales与pale(删除尾) pacb与pab(删除中间) pale与ale(删除头)

for (int i = 0, j = 0; i < len2; j++) {

if (count > 1) return false;

if (first.charAt(j) != second.charAt(i)) {

//不相同只移动长字符串的指针，把for改成while循环应该好些

count++; //不对应的地方

continue;

}

i++;

}

}

return true;

}

}