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@ -296,11 +296,15 @@ comments: true
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/* 回溯 */
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fn backtrack(choices: &[i32], state: i32, n: i32, res: &mut [i32]) {
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// 当爬到第 n 阶时,方案数量加 1
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if state == n { res[0] = res[0] + 1; }
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if state == n {
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res[0] = res[0] + 1;
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}
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// 遍历所有选择
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for &choice in choices {
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// 剪枝:不允许越过第 n 阶
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if state + choice > n { continue; }
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if state + choice > n {
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continue;
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}
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// 尝试:做出选择,更新状态
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backtrack(choices, state + choice, n, res);
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// 回退
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@ -309,7 +313,7 @@ comments: true
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/* 爬楼梯:回溯 */
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fn climbing_stairs_backtrack(n: usize) -> i32 {
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let choices = vec![ 1, 2 ]; // 可选择向上爬 1 阶或 2 阶
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let choices = vec![1, 2]; // 可选择向上爬 1 阶或 2 阶
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let state = 0; // 从第 0 阶开始爬
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let mut res = Vec::new();
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res.push(0); // 使用 res[0] 记录方案数量
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@ -592,7 +596,9 @@ $$
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/* 搜索 */
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fn dfs(i: usize) -> i32 {
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// 已知 dp[1] 和 dp[2] ,返回之
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if i == 1 || i == 2 { return i as i32; }
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if i == 1 || i == 2 {
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return i as i32;
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}
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// dp[i] = dp[i-1] + dp[i-2]
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let count = dfs(i - 1) + dfs(i - 2);
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count
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@ -600,7 +606,7 @@ $$
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/* 爬楼梯:搜索 */
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fn climbing_stairs_dfs(n: usize) -> i32 {
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dfs(n)
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dfs(n)
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}
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```
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@ -908,9 +914,13 @@ $$
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/* 记忆化搜索 */
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fn dfs(i: usize, mem: &mut [i32]) -> i32 {
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// 已知 dp[1] 和 dp[2] ,返回之
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if i == 1 || i == 2 { return i as i32; }
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if i == 1 || i == 2 {
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return i as i32;
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}
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// 若存在记录 dp[i] ,则直接返回之
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if mem[i] != -1 { return mem[i]; }
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if mem[i] != -1 {
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return mem[i];
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}
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// dp[i] = dp[i-1] + dp[i-2]
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let count = dfs(i - 1, mem) + dfs(i - 2, mem);
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// 记录 dp[i]
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@ -1186,7 +1196,9 @@ $$
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/* 爬楼梯:动态规划 */
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fn climbing_stairs_dp(n: usize) -> i32 {
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// 已知 dp[1] 和 dp[2] ,返回之
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if n == 1 || n == 2 { return n as i32; }
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if n == 1 || n == 2 {
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return n as i32;
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}
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// 初始化 dp 表,用于存储子问题的解
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let mut dp = vec![-1; n + 1];
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// 初始状态:预设最小子问题的解
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@ -1420,7 +1432,9 @@ $$
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```rust title="climbing_stairs_dp.rs"
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/* 爬楼梯:空间优化后的动态规划 */
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fn climbing_stairs_dp_comp(n: usize) -> i32 {
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if n == 1 || n == 2 { return n as i32; }
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if n == 1 || n == 2 {
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return n as i32;
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}
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let (mut a, mut b) = (1, 2);
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for _ in 3..=n {
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let tmp = b;
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