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130 lines
3.2 KiB
130 lines
3.2 KiB
8 months ago
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// File: edit_distance.go
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// Created Time: 2023-07-23
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// Author: Reanon (793584285@qq.com)
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package chapter_dynamic_programming
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/* 編輯距離:暴力搜尋 */
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func editDistanceDFS(s string, t string, i int, j int) int {
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// 若 s 和 t 都為空,則返回 0
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if i == 0 && j == 0 {
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return 0
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}
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// 若 s 為空,則返回 t 長度
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if i == 0 {
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return j
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}
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// 若 t 為空,則返回 s 長度
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if j == 0 {
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return i
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}
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// 若兩字元相等,則直接跳過此兩字元
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if s[i-1] == t[j-1] {
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return editDistanceDFS(s, t, i-1, j-1)
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}
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// 最少編輯步數 = 插入、刪除、替換這三種操作的最少編輯步數 + 1
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insert := editDistanceDFS(s, t, i, j-1)
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deleted := editDistanceDFS(s, t, i-1, j)
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replace := editDistanceDFS(s, t, i-1, j-1)
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// 返回最少編輯步數
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return MinInt(MinInt(insert, deleted), replace) + 1
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}
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/* 編輯距離:記憶化搜尋 */
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func editDistanceDFSMem(s string, t string, mem [][]int, i int, j int) int {
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// 若 s 和 t 都為空,則返回 0
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if i == 0 && j == 0 {
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return 0
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}
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// 若 s 為空,則返回 t 長度
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if i == 0 {
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return j
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}
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// 若 t 為空,則返回 s 長度
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if j == 0 {
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return i
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}
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// 若已有記錄,則直接返回之
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if mem[i][j] != -1 {
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return mem[i][j]
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}
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// 若兩字元相等,則直接跳過此兩字元
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if s[i-1] == t[j-1] {
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return editDistanceDFSMem(s, t, mem, i-1, j-1)
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}
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// 最少編輯步數 = 插入、刪除、替換這三種操作的最少編輯步數 + 1
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insert := editDistanceDFSMem(s, t, mem, i, j-1)
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deleted := editDistanceDFSMem(s, t, mem, i-1, j)
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replace := editDistanceDFSMem(s, t, mem, i-1, j-1)
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// 記錄並返回最少編輯步數
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mem[i][j] = MinInt(MinInt(insert, deleted), replace) + 1
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return mem[i][j]
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}
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/* 編輯距離:動態規劃 */
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func editDistanceDP(s string, t string) int {
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n := len(s)
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m := len(t)
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dp := make([][]int, n+1)
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for i := 0; i <= n; i++ {
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dp[i] = make([]int, m+1)
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}
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// 狀態轉移:首行首列
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for i := 1; i <= n; i++ {
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dp[i][0] = i
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}
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for j := 1; j <= m; j++ {
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dp[0][j] = j
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}
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// 狀態轉移:其餘行和列
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for i := 1; i <= n; i++ {
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for j := 1; j <= m; j++ {
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if s[i-1] == t[j-1] {
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// 若兩字元相等,則直接跳過此兩字元
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dp[i][j] = dp[i-1][j-1]
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} else {
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// 最少編輯步數 = 插入、刪除、替換這三種操作的最少編輯步數 + 1
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dp[i][j] = MinInt(MinInt(dp[i][j-1], dp[i-1][j]), dp[i-1][j-1]) + 1
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}
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}
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}
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return dp[n][m]
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}
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/* 編輯距離:空間最佳化後的動態規劃 */
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func editDistanceDPComp(s string, t string) int {
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n := len(s)
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m := len(t)
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dp := make([]int, m+1)
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// 狀態轉移:首行
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for j := 1; j <= m; j++ {
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dp[j] = j
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}
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// 狀態轉移:其餘行
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for i := 1; i <= n; i++ {
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// 狀態轉移:首列
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leftUp := dp[0] // 暫存 dp[i-1, j-1]
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dp[0] = i
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// 狀態轉移:其餘列
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for j := 1; j <= m; j++ {
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temp := dp[j]
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if s[i-1] == t[j-1] {
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// 若兩字元相等,則直接跳過此兩字元
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dp[j] = leftUp
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} else {
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// 最少編輯步數 = 插入、刪除、替換這三種操作的最少編輯步數 + 1
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dp[j] = MinInt(MinInt(dp[j-1], dp[j]), leftUp) + 1
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}
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leftUp = temp // 更新為下一輪的 dp[i-1, j-1]
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}
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}
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return dp[m]
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}
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func MinInt(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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