标签:assets 遍历 lse car 解题思路 lang 第一个字符 组成 解释
编写一个函数来查找字符串数组中的最长公共前缀。
如果不存在公共前缀,返回空字符串 ""
。
示例 1:
输入:strs = ["flower","flow","flight"]
输出:"fl"
示例 2:
输入:strs = ["dog","racecar","car"]
输出:""
解释:输入不存在公共前缀。
提示:
0 <= strs.length <= 200
0 <= strs[i].length <= 200
strs[i]
仅由小写英文字母组成如下图
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];
}
public String longestCommonPrefix(String[] strs) {
if (strs == null || strs.length == 0) {
return "";
} else {
return longestCommonPrefix(strs, 0, strs.length - 1);
}
}
public String longestCommonPrefix(String[] strs, int start, int end) {
if (start == end) {
return strs[start];
} else {
int mid = (end - start) / 2 + start;
String lcpLeft = longestCommonPrefix(strs, start, mid);
String lcpRight = longestCommonPrefix(strs, mid + 1, end);
return commonPrefix(lcpLeft, lcpRight);
}
}
public String commonPrefix(String lcpLeft, String lcpRight) {
int minLength = Math.min(lcpLeft.length(), lcpRight.length());
for (int i = 0; i < minLength; i++) {
if (lcpLeft.charAt(i) != lcpRight.charAt(i)) {
return lcpLeft.substring(0, i);
}
}
return lcpLeft.substring(0, minLength);
}
public String longestCommonPrefix(String[] strs) {
if (strs == null || strs.length == 0) {
return "";
}
int minLength = Integer.MAX_VALUE;
for (String str : strs) {
minLength = Math.min(minLength, str.length());
}
int low = 0, high = minLength;
// 二分法找最大的公共前缀
while (low < high) {
int mid = (high - low + 1) / 2 + low;
if (isCommonPrefix(strs, mid)) {
// 如果mid处是相同的, 则往右找
low = mid;
} else {
// 如果不是相同的, 则往左找
high = mid - 1;
}
}
return strs[0].substring(0, low);
}
public boolean isCommonPrefix(String[] strs, int length) {
// str0 代表第一个字符串
String str0 = strs[0].substring(0, length);
int count = strs.length;
// 遍历后面的每一个字符串
for (int i = 1; i < count; i++) {
String str = strs[i];
// 从字符串的首个字符逐个向后扫描, 看是否和第一个字符串的字符是相同的。
for (int j = 0; j < length; j++) {
if (str0.charAt(j) != str.charAt(j)) {
return false;
}
}
}
return true;
}
Easy | LeetCode 14. 最长公共前缀 | 二分法
标签:assets 遍历 lse car 解题思路 lang 第一个字符 组成 解释
原文地址:https://www.cnblogs.com/chenrj97/p/14456011.html