leetcode - 最长公共前缀
前端路桥
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·
2021-01-16 10:37
题意
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
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
出处
链接:https://leetcode-cn.com/problems/longest-common-prefix
思路
这里我们需要拿字符串数组 strs 中的第一个字符串 str 进行遍历,然后如果说截取到的字符串 cur 满足题意就把它赋值给 res,不满足就退出循环。这里有一点就是你要单独判断数组为 0 的情况, 然后如果你是用 substr 函数截取的话,长度可以取等号的。当然你也可以不借助 JS 的一些 API,手动遍历,然后字符串相加求解。
代码
巧用 JSAPI
/**
* @param {string[]} strs
* @return {string}
*/
const longestCommonPrefix = function (strs) {
if (strs.length === 0) {
return '';
}
let cur = '';
let res = '';
for (let i = 0; i <= strs[0].length; i++) {
cur = strs[0].substr(0, i);
if (strs.every((str) => str.startsWith(cur))) {
res = cur;
} else {
break;
}
}
return res;
};
export default longestCommonPrefix;
传统的解法
/**
* @param {string[]} strs
* @return {string}
*/
const longestCommonPrefix = function (strs) {
let res = '';
if (strs.length !== 0) {
for (let i = 0; i < strs[0].length; i++) {
if (strs.every((str) => str[i] === strs[0][i])) {
res += strs[0][i];
} else {
break;
}
}
}
return res;
};
export default longestCommonPrefix;
测试
import longestCommonPrefix from '../../code/leetcode/14';
describe('test function longestCommonPrefix:', () => {
test('test case strs = []', () => {
const res = longestCommonPrefix([]);
expect(res).toBe('');
});
test('test case strs = ["flower","flow","flight"]', () => {
const res = longestCommonPrefix(['flower', 'flow', 'flight']);
expect(res).toBe('fl');
});
test('test case strs = ["dog","racecar","car"]', () => {
const res = longestCommonPrefix(['dog', 'racecar', 'car']);
expect(res).toBe('');
});
test('test case strs = ["flower","flower","flower","flower"]', () => {
const res = longestCommonPrefix(['flower', 'flower', 'flower', 'flower']);
expect(res).toBe('flower');
});
test('test case strs = ["a"]', () => {
const res = longestCommonPrefix(['a']);
expect(res).toBe('a');
});
});
说明
本文首发于 GitHub 仓库https://github.com/ataola/coding
,线上阅读地址:https://zhengjiangtao.cn/coding/
,转载请注明出处,谢谢!
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