由于该题是套题中的第五题，所以我全都列出来，描述就只写本题的了
1.只允许交易一次：121. 买卖股票的最佳时机
2.无交易次数上限：122. 买卖股票的最佳时机 II
3.只允许交易两次：123.买卖股票的最佳时机 III
4.有交易次数上限：188. 买卖股票的最佳时机 IV
5.每次交易后休息：309. 最佳买卖股票时机含冷冻期
6.每次交易要费用：714. 买卖股票的最佳时机含手续费
题目描述：
给定一个整数数组，其中第 i 个元素代表了第 i 天的股票价格 。
设计一个算法计算出最大利润。在满足以下约束条件下，你可以尽可能地完成更多的交易（多次买卖一支股票）:
你不能同时参与多笔交易（你必须在再次购买前出售掉之前的股票）。
卖出股票后，你无法在第二天买入股票 (即冷冻期为 1 天)。
今日学习：
1.动规动规动动规
题解：
首先参考：模板动规：一穿六
状态转移图:
状态转移方程：
base case：
dp[-1][k][0] = dp[i][0][0] = 0
dp[-1][k][1] = dp[i][0][1] = -infinity
状态转移方程：
dp[i][k][0] = max(dp[i-1][k][0], dp[i-1][k][1] + prices[i])
dp[i][k][1] = max(dp[i-1][k][1], dp[i-1][k-1][0] - prices[i])
k为1或不限制时，都可以省略k的存在，当k有限制时需要循环k进行枚举。
题解1只允许交易一次：
/**
* @param {number[]} prices
* @return {number}
*/
//暴力法
var maxProfit = function(prices) {
let res = 0
for(let i = 0; i < prices.length - 1; i++) {
let max = Math.max(...prices.slice(i + 1))
res = Math.max(res, max - prices[i])
}
return res
};
//动规
var maxProfit = function(prices) {
let min = Number.MAX_VALUE
let res = 0
for(let i = 0; i < prices.length; i++) {
if(prices[i] < min) {
min = prices[i]
}
res = Math.max(res, prices[i] - min)
}
return res
}
//模板动规
var maxProfit = function(prices) {
let n = prices.length
let dp_i_0 = 0, dp_i_1 = -prices[0]
for(let i = 0; i < n; i++) {
dp_i_0 = Math.max(dp_i_0, dp_i_1 + prices[i])
dp_i_1 = Math.max(dp_i_1, -prices[i])
}
return dp_i_0
}
题解2无交易次数上限：
/**
* @param {number[]} prices
* @return {number}
*/
//峰谷
var maxProfit = function(prices) {
let i = 0
let valley = peak = prices[0]
let max = 0
let len = prices.length
while(i < len - 1) {
while(i < len - 1 && prices[i] >= prices[i + 1]) {
i++
}
valley = prices[i]
while(i < len - 1 && prices[i] <= prices[i + 1]) {
i++
}
peak = prices[i]
max += peak - valley
}
return max
};
//连续计算
var maxProfit = function(prices) {
let max = 0
for(let i = 1; i < prices.length; i++) {
if(prices[i] > prices[i - 1]) {
max += prices[i] - prices[i - 1]
}
}
return max
}
//模板动规
var maxProfit = function(prices) {
let n = prices.length
let dp_i_0 = 0, dp_i_1 = -prices[0]
for(let i = 0; i < n; i++) {
dp_i_0 = Math.max(dp_i_0, dp_i_1 + prices[i])
dp_i_1 = Math.max(dp_i_1, dp_i_0 - prices[i])
}
return dp_i_0
}
题解3只允许交易两次：
/**
* @param {number[]} prices
* @return {number}
*/
//暴力法
var maxProfit = function(prices) {
let max = 0
for(let i = 0; i < prices.length; i++) {
let p1 = prices.slice(0, i + 1)
let p2 = prices.slice(i + 1)
max = Math.max(max, maxProfitOnce(p1) + maxProfitOnce(p2))
}
return max
};
var maxProfitOnce = function(prices) {
let min = Number.MAX_VALUE
let res = 0
for(let i = 0; i < prices.length; i++) {
if(prices[i] < min) {
min = prices[i]
}
res = Math.max(res, prices[i] - min)
}
return res
}
//动规
var maxProfit = function(prices) {
let n = prices.length;
if(n == 0){
return 0;
}
let maxTime = 2;
let dp = Array.from(new Array(n),() => new Array(maxTime+1));
for(let i = 0;i < n;i++){
for(let r = 0;r <= maxTime;r++){
dp[i][r] = new Array(2).fill(0);
}
}
for(let i = 0;i < n;i++){
for(let k = maxTime;k >= 1;k--){
if(i == 0){
dp[i][k][0] = 0;
dp[i][k][1] = -prices[i];
continue;
}
dp[i][k][0] = Math.max(dp[i-1][k][0],dp[i-1][k][1] + prices[i]);
dp[i][k][1] = Math.max(dp[i-1][k][1],dp[i-1][k-1][0] - prices[i]);
}
}
return dp[n-1][maxTime][0];
};
//模板动规
var maxProfit = function(prices) {
let n = prices.length
//第一次交易：没持有（没买）利润是0，持有（买了）利润是-prices[0]
let dp_i_1_0 = 0, dp_i_1_1 = -prices[0]
//初始第二次交易和第一次相同
let dp_i_2_0 = 0, dp_i_2_1 = -prices[0]
for(let i = 1; i < n; i++) {
dp_i_1_0 = Math.max(dp_i_1_0, dp_i_1_1 + prices[i])
dp_i_1_1 = Math.max(dp_i_1_1, -prices[i])
dp_i_2_0 = Math.max(dp_i_2_0, dp_i_2_1 + prices[i])
dp_i_2_1 = Math.max(dp_i_2_1, dp_i_1_0 - prices[i])
}
return dp_i_2_0
};
题解4有交易次数上限：
/**
* @param {number} k
* @param {number[]} prices
* @return {number}
*/
var maxProfit = function(k, prices) {
let n = prices.length
if (k > n / 2) {return maxProfitINF(prices)}
let dp = Array.from(
{length: n+1},() => Array.from(
{length: k + 1},() => new Array(2))
)
for(let K = 0; K <= k; K++) {
dp[0][K][0] = 0;
dp[0][K][1] = -prices[0];
}
for(let i = 1; i < n + 1; i++) {
dp[i][0][0] = 0;
dp[i][0][1] = -prices[0];
for(let j = 1; j < k + 1; j++) {
dp[i][j][0] = Math.max(dp[i-1][j][0],dp[i-1][j][1] + prices[i-1]);
dp[i][j][1] = Math.max(dp[i-1][j][1],dp[i-1][j-1][0]-prices[i-1]);
}
}
return dp[n][k][0];
};
var maxProfitINF = function(prices) {
let n = prices.length
let dp_i_0 = 0, dp_i_1 = -prices[0]
for(let i = 0; i < n; i++) {
dp_i_0 = Math.max(dp_i_0, dp_i_1 + prices[i])
dp_i_1 = Math.max(dp_i_1, dp_i_0 - prices[i])
}
return dp_i_0
}
题解5每次交易后休息：
/**
* @param {number[]} prices
* @return {number}
*/
var maxProfit = function(prices) {
let n = prices.length
let dp_i_0 = 0, dp_i_1 = -prices[0]
let dp_pre_0 = 0
for(let i = 0; i < n; i++) {
let temp = dp_i_0
dp_i_0 = Math.max(dp_i_0, dp_i_1 + prices[i])
dp_i_1 = Math.max(dp_i_1, dp_pre_0 - prices[i])
dp_pre_0 = temp
}
return dp_i_0
};
题解6每次交易要费用：
/**
* @param {number[]} prices
* @param {number} fee
* @return {number}
*/
var maxProfit = function(prices, fee) {
let n = prices.length
let dp_i_0 = 0, dp_i_1 = -prices[0]
for(let i = 0; i < n; i++) {
dp_i_0 = Math.max(dp_i_0, dp_i_1 + prices[i] - fee)
dp_i_1 = Math.max(dp_i_1, dp_i_0 - prices[i])
}
return dp_i_0
};