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Commit e5ab214

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Merge branch 'master' of github.com:jinbudaily/leetcode-master
合并远程代码
2 parents 28b5878 + f90e8a2 commit e5ab214

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‎problems/0123.买卖股票的最佳时机III.md‎

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@@ -413,7 +413,52 @@ function maxProfit(prices: number[]): number {
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};
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```
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Rust:
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> 版本一
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```rust
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impl Solution {
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pub fn max_profit(prices: Vec<i32>) -> i32 {
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/*
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* 定义 5 种状态:
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* 0: 没有操作, 1: 第一次买入, 2: 第一次卖出, 3: 第二次买入, 4: 第二次卖出
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*/
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let mut dp = vec![vec![0; 5]; prices.len()];
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dp[0][1] = -prices[0];
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dp[0][3] = -prices[0];
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for (i, &p) in prices.iter().enumerate().skip(1) {
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// 不操作
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// dp[i][0] = dp[i - 1][0];
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dp[i][1] = dp[i - 1][1].max(-p);
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dp[i][2] = dp[i - 1][2].max(dp[i - 1][1] + p);
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dp[i][3] = dp[i - 1][3].max(dp[i - 1][2] - p);
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dp[i][4] = dp[i - 1][4].max(dp[i - 1][3] + p);
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}
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dp[prices.len() - 1][4]
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}
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}
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```
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> 版本二(绕)
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```rust
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impl Solution {
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pub fn max_profit(prices: Vec<i32>) -> i32 {
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let (mut one_buy, mut one_sale, mut two_buy, mut two_sale) = (-prices[0], 0, -prices[0], 0);
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for p in prices {
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one_buy = one_buy.max(-p);
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one_sale = one_sale.max(p + one_buy);
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two_buy = two_buy.max(one_sale - p);
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two_sale = two_sale.max(two_buy + p);
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}
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two_sale
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}
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}
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```
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<p align="center">

‎problems/0188.买卖股票的最佳时机IV.md‎

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@@ -474,6 +474,55 @@ function maxProfit(k: number, prices: number[]): number {
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};
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```
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Rust:
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```rust
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impl Solution {
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pub fn max_profit(k: i32, prices: Vec<i32>) -> i32 {
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let mut dp = vec![vec![0; 2 * k as usize + 1]; prices.len()];
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for v in dp[0].iter_mut().skip(1).step_by(2) {
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*v = -prices[0];
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}
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for (i, &p) in prices.iter().enumerate().skip(1) {
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for j in (0..2 * k as usize - 1).step_by(2) {
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dp[i][j + 1] = dp[i - 1][j + 1].max(dp[i - 1][j] - p);
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dp[i][j + 2] = dp[i - 1][j + 2].max(dp[i - 1][j + 1] + p);
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}
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}
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dp[prices.len() - 1][2 * k as usize]
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}
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}
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```
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空间优化:
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```rust
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impl Solution {
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pub fn max_profit(k: i32, prices: Vec<i32>) -> i32 {
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let mut dp = vec![0; 2 * k as usize + 1];
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for v in dp.iter_mut().skip(1).step_by(2) {
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*v = -prices[0];
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}
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for p in prices {
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for i in 1..=2 * k as usize {
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if i % 2 == 1 {
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// 买入
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dp[i] = dp[i].max(dp[i - 1] - p);
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continue;
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}
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// 卖出
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dp[i] = dp[i].max(dp[i - 1] + p);
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}
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}
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dp[2 * k as usize]
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}
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}
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```
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<p align="center">

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