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42 lines (41 loc) · 1.51 KB
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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;
}
}