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0204Leetcode.js
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0204Leetcode.js
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var longestCommonPrefix = function(strs) {
'use strict';
if (strs === undefined || strs.length === 0) { return ''; }
//reduce 는 value 가 하나일때 prev, next 모두가 같은 value 이다.
// ex) [1].reduce((prev, next)=> console.log(prev, next)) ==> prev = 1, next = 1
return strs.reduce((prev, next) => {
let i = 0;
console.log(prev)
// If there is more than one value, we compare. (there should be prev(first value) and next(second value) and they should be equal
while (prev[i] && next[i] && prev[i] === next[i]) {
// prev 와 next 가 같음으로 flower 워의 모든 index 즉 i는 5까지 쭉 간다
i++;
}
return prev.slice(0, i);
});
};
console.log(longestCommonPrefix(["flower"]))
// console.log(longestCommonPrefix(["flower", "abcd", "qzp"]))
// console.log(longestCommonPrefix(["flower", "florida", "Flow"]))
// console.log(longestCommonPrefix(["flower", "florida", "floW"]))
var longestCommonPrefix2 = function(strs) {
if (!strs.length) return ''; // If there is nothing in array. Minimum
let prefix = '';
// 오래 볼것도 없다. value 들중 가장 짧은 단어만 가져온다.
let maxPrefixLength = Math.min(...strs.map(str => str.length));
// 짧은 단어의 길이만큼만 돈다.
for (let i = 0; i < maxPrefixLength; i++) {
let char = strs[0][i];
// iterating through characters of each value with i
if (strs.every(str => str[i] === char)) {
prefix += char;
} else {
break;
}
};
return prefix;
};
// console.log(longestCommonPrefix2(["flower"]))
// console.log(longestCommonPrefix(["flower", "florida", "floW"]))
// console.log(longestCommonPrefix(["flower", "florida", "Flow"]))
// console.log(longestCommonPrefix(["flower", "florida", "floW"]))