https://leetcode.com/problems/longest-word-in-dictionary/description/
题目
Given a list of strings words representing an English Dictionary, find the longest word in words that can be built one character at a time by other words in words. If there is more than one possible answer, return the longest word with the smallest lexicographical order. If there is no answer, return the empty string.
Example 1:
Input:
words = ["w","wo","wor","worl", "world"]
Output: "world"
Explanation:
The word "world" can be built one character at a time by "w", "wo", "wor", and "worl".
Example 2:
Input:
words = ["a", "banana", "app", "appl", "ap", "apply", "apple"]
Output: "apple"
Explanation:
Both "apply" and "apple" can be built from other words in the dictionary. However, "apple" is lexicographically smaller than "apply".
Note:
- All the strings in the input will only contain lowercase letters.
- The length of words will be in the range [1, 1000].
- The length of words[i] will be in the range [1, 30].
题目翻译
题目解析
参考答案
package main
import (
"sort"
"fmt"
)
type SortByString []string
func (s SortByString) Len() int {
return len(s)
}
func (s SortByString) Swap(i, j int) {
s[i], s[j] = s[j], s[i]
}
func (s SortByString) Less(i, j int) bool {
return s[i] < s[j]
}
func longestWord(words []string) string {
lWord := ""
m := make(map[string]bool)
fmt.Println(words)
sort.Sort(SortByString(words))
fmt.Println(words)
for _, v := range words {
bSuccess := false
if len(v) == 1 {
m[v] = true
bSuccess = true
} else {
prefix := v[:len(v)-1]
fmt.Println("v = ", v, ", prefix = ", prefix)
if _, ok := m[prefix]; ok {
m[v] = true
bSuccess = true
}
}
if bSuccess {
if lWord == "" {
lWord = v
} else {
if (len(v) > len(lWord)) || (len(v) == len(lWord) && v < lWord) {
lWord = v
}
}
}
}
return lWord
}
func main() {
words := []string{
"w", "wo", "wor", "worl", "world",
}
ret := longestWord(words)
fmt.Println("ret = ", ret)
}
