algorithm/hello-algo
1
0
Fork 0
mirror of https://github.com/krahets/hello-algo.git synced 2025-07-22 15:53:50 +08:00
Code Issues Packages Projects Releases Wiki Activity

build

Browse Source
This commit is contained in:
krahets
2024-03-21 04:22:07 +08:00
parent 35a07170c0
commit cfdb743939
52 changed files with 292 additions and 290 deletions
Download Patch File Download Diff File Expand all files Collapse all files

6
docs/chapter_dynamic_programming/dp_problem_features.md
View File

@ -164,7 +164,7 @@ $$
dp[1] = cost[1]
dp[2] = cost[2]
// 状态转移：从较小子问题逐步求解较大子问题
for i in stride(from: 3, through: n, by: 1) {
for i in 3 ... n {
dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
}
return dp[n]
@ -421,7 +421,7 @@ $$
return cost[n]
}
var (a, b) = (cost[1], cost[2])
for i in stride(from: 3, through: n, by: 1) {
for i in 3 ... n {
(a, b) = (b, min(a, b) + cost[i])
}
return b
@ -721,7 +721,7 @@ $$
dp[2][1] = 0
dp[2][2] = 1
// 状态转移：从较小子问题逐步求解较大子问题
for i in stride(from: 3, through: n, by: 1) {
for i in 3 ... n {
dp[i][1] = dp[i - 1][2]
dp[i][2] = dp[i - 2][1] + dp[i - 2][2]
}

14
docs/chapter_dynamic_programming/dp_solution_pipeline.md
View File

@ -861,16 +861,16 @@ $$
var dp = Array(repeating: Array(repeating: 0, count: m), count: n)
dp[0][0] = grid[0][0]
// 状态转移：首行
for j in stride(from: 1, to: m, by: 1) {
for j in 1 ..< m {
dp[0][j] = dp[0][j - 1] + grid[0][j]
}
// 状态转移：首列
for i in stride(from: 1, to: n, by: 1) {
for i in 1 ..< n {
dp[i][0] = dp[i - 1][0] + grid[i][0]
}
// 状态转移：其余行和列
for i in stride(from: 1, to: n, by: 1) {
for j in stride(from: 1, to: m, by: 1) {
for i in 1 ..< n {
for j in 1 ..< m {
dp[i][j] = min(dp[i][j - 1], dp[i - 1][j]) + grid[i][j]
}
}
@ -1244,15 +1244,15 @@ $$
var dp = Array(repeating: 0, count: m)
// 状态转移：首行
dp[0] = grid[0][0]
for j in stride(from: 1, to: m, by: 1) {
for j in 1 ..< m {
dp[j] = dp[j - 1] + grid[0][j]
}
// 状态转移：其余行
for i in stride(from: 1, to: n, by: 1) {
for i in 1 ..< n {
// 状态转移：首列
dp[0] = dp[0] + grid[i][0]
// 状态转移：其余列
for j in stride(from: 1, to: m, by: 1) {
for j in 1 ..< m {
dp[j] = min(dp[j - 1], dp[j]) + grid[i][j]
}
}

14
docs/chapter_dynamic_programming/edit_distance_problem.md
View File

@ -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

4
docs/chapter_dynamic_programming/intro_to_dynamic_programming.md
View File

@ -1126,7 +1126,7 @@ $$
dp[1] = 1
dp[2] = 2
// 状态转移：从较小子问题逐步求解较大子问题
for i in stride(from: 3, through: n, by: 1) {
for i in 3 ... n {
dp[i] = dp[i - 1] + dp[i - 2]
}
return dp[n]
@ -1370,7 +1370,7 @@ $$
}
var a = 1
var b = 2
for _ in stride(from: 3, through: n, by: 1) {
for _ in 3 ... n {
(a, b) = (b, a + b)
}
return b

8
docs/chapter_dynamic_programming/knapsack_problem.md
View File

@ -802,8 +802,8 @@ $$
// 初始化 dp 表
var dp = Array(repeating: Array(repeating: 0, count: cap + 1), count: n + 1)
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for c in stride(from: 1, through: cap, by: 1) {
for i in 1 ... n {
for c in 1 ... cap {
if wgt[i - 1] > c {
// 若超过背包容量，则不选物品 i
dp[i][c] = dp[i - 1][c]
@ -1191,9 +1191,9 @@ $$
// 初始化 dp 表
var dp = Array(repeating: 0, count: cap + 1)
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for i in 1 ... n {
// 倒序遍历
for c in stride(from: cap, through: 1, by: -1) {
for c in (1 ... cap).reversed() {
if wgt[i - 1] <= c {
// 不选和选物品 i 这两种方案的较大值
dp[c] = max(dp[c], dp[c - wgt[i - 1]] + val[i - 1])

28
docs/chapter_dynamic_programming/unbounded_knapsack_problem.md
View File

@ -166,8 +166,8 @@ $$
// 初始化 dp 表
var dp = Array(repeating: Array(repeating: 0, count: cap + 1), count: n + 1)
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for c in stride(from: 1, through: cap, by: 1) {
for i in 1 ... n {
for c in 1 ... cap {
if wgt[i - 1] > c {
// 若超过背包容量，则不选物品 i
dp[i][c] = dp[i - 1][c]
@ -506,8 +506,8 @@ $$
// 初始化 dp 表
var dp = Array(repeating: 0, count: cap + 1)
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for c in stride(from: 1, through: cap, by: 1) {
for i in 1 ... n {
for c in 1 ... cap {
if wgt[i - 1] > c {
// 若超过背包容量，则不选物品 i
dp[c] = dp[c]
@ -882,12 +882,12 @@ $$
// 初始化 dp 表
var dp = Array(repeating: Array(repeating: 0, count: amt + 1), count: n + 1)
// 状态转移：首行首列
for a in stride(from: 1, through: amt, by: 1) {
for a in 1 ... amt {
dp[0][a] = MAX
}
// 状态转移：其余行和列
for i in stride(from: 1, through: n, by: 1) {
for a in stride(from: 1, through: amt, by: 1) {
for i in 1 ... n {
for a in 1 ... amt {
if coins[i - 1] > a {
// 若超过目标金额，则不选硬币 i
dp[i][a] = dp[i - 1][a]
@ -1300,8 +1300,8 @@ $$
var dp = Array(repeating: MAX, count: amt + 1)
dp[0] = 0
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for a in stride(from: 1, through: amt, by: 1) {
for i in 1 ... n {
for a in 1 ... amt {
if coins[i - 1] > a {
// 若超过目标金额，则不选硬币 i
dp[a] = dp[a]
@ -1660,12 +1660,12 @@ $$
// 初始化 dp 表
var dp = Array(repeating: Array(repeating: 0, count: amt + 1), count: n + 1)
// 初始化首列
for i in stride(from: 0, through: n, by: 1) {
for i in 0 ... n {
dp[i][0] = 1
}
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for a in stride(from: 1, through: amt, by: 1) {
for i in 1 ... n {
for a in 1 ... amt {
if coins[i - 1] > a {
// 若超过目标金额，则不选硬币 i
dp[i][a] = dp[i - 1][a]
@ -2002,8 +2002,8 @@ $$
var dp = Array(repeating: 0, count: amt + 1)
dp[0] = 1
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for a in stride(from: 1, through: amt, by: 1) {
for i in 1 ... n {
for a in 1 ... amt {
if coins[i - 1] > a {
// 若超过目标金额，则不选硬币 i
dp[a] = dp[a]