题目:
字符串有三种编辑操作:插入一个英文字符、删除一个英文字符或者替换一个英文字符。 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。
示例 1:
输入:
first = "pale"
second = "ple"
输出: True
示例 2:
输入:
first = "pales"
second = "pal"
输出: False
代码实现:
class Solution {标签:yyds,return,金典,second,index2,false,index1,LeetCode,first From: https://blog.51cto.com/u_13321676/5903574
public boolean oneEditAway(String first, String second) {
int m = first.length(), n = second.length();
if (n - m == 1) {
return oneInsert(first, second);
} else if (m - n == 1) {
return oneInsert(second, first);
} else if (m == n) {
boolean foundDifference = false;
for (int i = 0; i < m; i++) {
if (first.charAt(i) != second.charAt(i)) {
if (!foundDifference) {
foundDifference = true;
} else {
return false;
}
}
}
return true;
} else {
return false;
}
}
public boolean oneInsert(String shorter, String longer) {
int length1 = shorter.length(), length2 = longer.length();
int index1 = 0, index2 = 0;
while (index1 < length1 && index2 < length2) {
if (shorter.charAt(index1) == longer.charAt(index2)) {
index1++;
}
index2++;
if (index2 - index1 > 1) {
return false;
}
}
return true;
}
}