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krahets
2024-03-21 04:22:07 +08:00
parent 35a07170c0
commit cfdb743939
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14
docs/chapter_dynamic_programming/edit_distance_problem.md
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@ -234,15 +234,15 @@ $$
let m = t.utf8CString.count
var dp = Array(repeating: Array(repeating: 0, count: m + 1), count: n + 1)
// 状态转移：首行首列
for i in stride(from: 1, through: n, by: 1) {
for i in 1 ... n {
dp[i][0] = i
}
for j in stride(from: 1, through: m, by: 1) {
for j in 1 ... m {
dp[0][j] = j
}
// 状态转移：其余行和列
for i in stride(from: 1, through: n, by: 1) {
for j in stride(from: 1, through: m, by: 1) {
for i in 1 ... n {
for j in 1 ... m {
if s.utf8CString[i - 1] == t.utf8CString[j - 1] {
// 若两字符相等，则直接跳过此两字符
dp[i][j] = dp[i - 1][j - 1]
@ -681,16 +681,16 @@ $$
let m = t.utf8CString.count
var dp = Array(repeating: 0, count: m + 1)
// 状态转移：首行
for j in stride(from: 1, through: m, by: 1) {
for j in 1 ... m {
dp[j] = j
}
// 状态转移：其余行
for i in stride(from: 1, through: n, by: 1) {
for i in 1 ... n {
// 状态转移：首列
var leftup = dp[0] // 暂存 dp[i-1, j-1]
dp[0] = i
// 状态转移：其余列
for j in stride(from: 1, through: m, by: 1) {
for j in 1 ... m {
let temp = dp[j]
if s.utf8CString[i - 1] == t.utf8CString[j - 1] {
// 若两字符相等，则直接跳过此两字符