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
};