Substring with Concatenation of
2018-09-22 本文已影响0人
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Substring with Concatenation of All Words
题目描述
You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.
Example 1:
Input:
s = "barfoothefoobarman",
words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.
Example 2:
Input:
s = "wordgoodstudentgoodword",
words = ["word","student"]
Output: []
思路:
- 题的特点是被匹配字串都是等长的。所以才能将子串的单词子串,易于用哈希O(1)查找。
- 匹配集合用哈希表计数。遍历子串及子串的单词子串,存在则加入已匹配哈希。
- 这些情况放弃当前子串匹配:单词子串匹配次数超出,单词子串不匹配。
细节:
- 输入条件检查: if s == "" || len(words) == 0{return nil}
- 哈希表put需要判断,所以多写了几行代码。
解:
package main
import "fmt"
func findSubstring(s string, words []string) []int {
if s == "" || len(words) == 0{return nil}
res := make([]int,0,len(s))
sSlice := []rune(s)
wordsMap := make(map[string]int, len(words))
lenWord := len(words[0])
lenSubstr := lenWord * len(words)
for _, elem := range words {
_, ok := wordsMap[elem]
if ok {
wordsMap[elem] = wordsMap[elem] + 1
} else {
wordsMap[elem] = 1
}
}
for i := 0; i <= len(s)-lenSubstr; i++ {
subMap := make(map[string]int,len(words))
j :=0
for ;j<len(words);j++ {
subStr := string(sSlice[i+j*lenWord:i+j*lenWord+lenWord])
if num,ok := wordsMap[subStr];ok {
if numSub,ok := subMap[subStr];ok{
subMap[subStr] = numSub+1
if numSub+1 > num {
break
}
}else {
subMap[subStr] = 1
}
} else {
break
}
}
if j == len(words) {
res = append(res,i)
}
}
if len(res) == 0 {return nil}
return res
}
func main() {
s := "aaaaaaaa"
words := []string{"aa","aa","aa"}
res := findSubstring(s,words)
fmt.Print(res)
}