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| 1 | +package leetcode |
| 2 | + |
| 3 | +import ( |
| 4 | + "sort" |
| 5 | +) |
| 6 | + |
| 7 | +func closeStrings(word1 string, word2 string) bool { |
| 8 | + if len(word1) != len(word2) { |
| 9 | + return false |
| 10 | + } |
| 11 | + freqWord1, freq1, freqList1, freqWord2, freq2, freqList2, flag := map[byte]int{}, []int{}, map[int][]byte{}, map[byte]int{}, []int{}, map[int][]byte{}, false |
| 12 | + for i := 0; i < len(word1); i++ { |
| 13 | + freqWord1[word1[i]]++ |
| 14 | + } |
| 15 | + for i := 0; i < len(word2); i++ { |
| 16 | + freqWord2[word2[i]]++ |
| 17 | + } |
| 18 | + freqTemp1 := map[int]int{} |
| 19 | + for k, v := range freqWord1 { |
| 20 | + freqTemp1[v]++ |
| 21 | + if list, ok := freqList1[v]; ok { |
| 22 | + list = append(list, k) |
| 23 | + freqList1[v] = list |
| 24 | + } else { |
| 25 | + list := []byte{} |
| 26 | + list = append(list, k) |
| 27 | + freqList1[v] = list |
| 28 | + } |
| 29 | + } |
| 30 | + for _, v := range freqTemp1 { |
| 31 | + freq1 = append(freq1, v) |
| 32 | + } |
| 33 | + freqTemp2 := map[int]int{} |
| 34 | + for k, v := range freqWord2 { |
| 35 | + freqTemp2[v]++ |
| 36 | + if list, ok := freqList2[v]; ok { |
| 37 | + list = append(list, k) |
| 38 | + freqList2[v] = list |
| 39 | + } else { |
| 40 | + list := []byte{} |
| 41 | + list = append(list, k) |
| 42 | + freqList2[v] = list |
| 43 | + } |
| 44 | + } |
| 45 | + for _, v := range freqTemp2 { |
| 46 | + freq2 = append(freq2, v) |
| 47 | + } |
| 48 | + if len(freq1) != len(freq2) { |
| 49 | + return false |
| 50 | + } |
| 51 | + sort.Ints(freq1) |
| 52 | + sort.Ints(freq2) |
| 53 | + for i := 0; i < len(freq1); i++ { |
| 54 | + if freq1[i] != freq2[i] { |
| 55 | + flag = true |
| 56 | + break |
| 57 | + } |
| 58 | + } |
| 59 | + if flag == true { |
| 60 | + return false |
| 61 | + } |
| 62 | + flag = false |
| 63 | + // 频次相同,再判断字母交换是否合法存在 |
| 64 | + for k, v := range freqWord1 { |
| 65 | + if list, ok := freqList2[v]; ok { |
| 66 | + for i := 0; i < len(list); i++ { |
| 67 | + if list[i] != k && list[i] != '0' { |
| 68 | + // 交换的字母不存在 |
| 69 | + if _, ok := freqWord1[list[i]]; !ok { |
| 70 | + flag = true |
| 71 | + break |
| 72 | + } else { |
| 73 | + // 交换的字母存在,重置这一位,代表这一个字母被交换了,下次不用它 |
| 74 | + list[i] = '0' |
| 75 | + } |
| 76 | + } |
| 77 | + } |
| 78 | + } else { |
| 79 | + // 出现频次个数相同,但是频次不同 |
| 80 | + flag = true |
| 81 | + break |
| 82 | + } |
| 83 | + } |
| 84 | + if flag == true { |
| 85 | + return false |
| 86 | + } |
| 87 | + return true |
| 88 | +} |
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