我正在尝试解决问题，并提出一个可以正确回答该问题的重复关系。代码：
private static int recurse( int[] prices, int index, int[][] dp ) {
if (index == prices.length - 1) {
return 0;
}
int profit = 0, min = Integer.MAX_VALUE;
int m = index;
for (; m < prices.length; m++) {
min = Math.min(min, prices[m]);//picking at min prices
int diff = prices[m] - min;//diff. for current min. picked stock profit
if (dp[m][index] != -1)
return dp[m][index];
if (diff > 0) {
diff += recurse(prices, m + 1, dp);
}
profit = Math.max(profit, diff - fees);
}
dp[m][index] = profit;
return profit;
}
上述解决方案在测试用例中返回TLE：
arr[] = [1036,2413,2776,825,2640,31,1560,2917,4282,783,3146,2600,1939,694,4284,3881,554,167,372,4620,3037,1175,1075,3845,4981,4495,3406,4228,2807,4774,4526,3914,2633,3762,1570,2334,616,1648,1914,2900,349,2428,4013,1964,4020,1882,629,240,2595,2902,3419,292,224,4437,4918,632,3701,3840,3996,2129,3345,3979,1954,781,1576,1084,3250,4517,3106,2133,309,4520,2225,4366,4628,1303,4373,1266,3181,558,3855,3447,4335,2115,4603,661,1715,3972,2846,342,686,787,273,2575,100,2860,3587,4236,3862,2238,3471,3123,431,4489,1551,596,4037,4986,594,2386,326,628,1363,2377,4986,3780,3853,2670,2852,3519,2998,4083,3392,2394,1083,3958,4082,1506,2322,2715,4901,2555,4097,3748,4717,3901,3329,4616,3334,2603,3705,631,3541,555,508,464,4495,4463,3616,31,2177,3307,1011,2759,751,1537,1000,292,3921,1442,2726,4677,792,82,2580,609,4758,3190,1958,913,955,1259,1634,4729,2672,1761,1467,2347,4295,2049,4708,1452,3411,1428,4078,2627,3785,2432,2916,492,1108,1691,972,3823,4086,2115,1925,1454,291,3266,300,2539,2681,2084,4633,1084,1061,1043,1304,2205,410,4332,2567,703,529,4273,3684,308,3164,4876,3108,4993,4555,1237,4753,549,2795,3426,819,2897,825,2514,3419,1854,3209,3766,2794,4117,4668,2162,1571,2446,1480,974,1090,3903,4655,4452,1451,2953,1241,842,1750,3847,3053,4395,4338,1493,1660,1569,3418,3029,4416,2056,2283,3392,2032,4354,803,4959,3630,2080,1553,873,4050,1986,2328,55,4602,1430,4238,4326,3382,4845,4968,1903,423,4717,2427,4618,2644,4541,380,3404,4880,2577,1640,189,2692,3788,818,4091,4730,611,1776,3594,4746,580,2083,4183,3355,3063,658,4532,3318,3902,556,2249,4653,2118,1529,4793,4935,4259,3542,1705,2839,1436,3918,564,3277,2988,2460,3213,4445,4238,1954,2213,1748,939,1149,1408,2408,1781,1618,1457,2123,3366,826,2094,16,1161,3337,1864,433,1303,4800,4667,4769,1026,3440,1072,4725,6,1263,4184,2728,1315,2091,3032,2071,2672,4557,1916,638,2133,2687,2408,1677,344,697,1699,8,480,655,2656,4983,455,1611,1726,692,392,1921,2555,3549,3740,3840,3062,3420,2428,1169,4570,389,3509,2169,3290,1680,1733,1765,2518,3260,3644,765,4521,269,2501,4014,1743,239,4908,1656,4433,3647,2612,4872,387,3091,4011,564,4421,810,3623,3451,4108,1428,475,3755,4484,3527,3062,4706,3424,2678,2411,4446,2556,4305,1305,646,1458,4471,1689,4556,3851,1245,1197,3785,1175,2904,302,2422,4302,2148,2338,4288,375,2824,1623,3717,1142,4254,192,783,1963,2225,1209,1746,3072,2737,4640,4919,3614,804,4029,1751,2360,3789,4445,2283,2769,2833,4452,2978,2809,4532,4365,2124,3541,2658,2902,4688,3980,1543,4041,1420,1452,1284,66,19,947,932,3244,3374,1910,2561,3466,4104,1667,589,3048,730,1770,1241,2270,4016,2835,604,4771,514,3854,3427,1875,2038,3067,3216,4732,3735,4440,2855,4958,4569,1685,3539,4589,3512,3143,898,3004,3072,2573,3163,2522,3927,330,3874,363,1900,1629,1156,4259,2747,3445,4513,2867,52,3870,1761,619,3308,4380,1101,2592,4852,4140,174,3997,4617,3500,3028,907,2355,759,374,2429,412,2132,3973,3583,3028,2070,2235,2659,1053,2558,753,1221,1185,2225,1593,3554,3703,332,2843,3349,3871,4389,6,2768,4382,902,417,191,2107,2838,4958,3905,4966,3937,1105,4150,2682,3396,818,2297,2077,2032,3340,2478,127,4379,954,2593,3454,1230,2308,3694,2179,4134,653,3808,4043,2069,660,4515,4189,4876,1784,4166,342,1766,3305,1980,1909,4115,4115,1461,2061,838,3112,122,656,4856,4822,3468,2111,2700,4124,4663,2948,3029,4182,3847,4760,1323,1505,308,128,874,583,2671,1315,747,2682,2841,67,2712,2703,4471,2952,3081,464,655,57,1460,1395,682,2447,2590,4624,1578,64,4060,2975,1236,831,3313,1432,2589,3777,1868,1720,45,3311,4532,2672,454,752,4839,4717,748,4323,2999,3491,631,1407,1453,4611,4263,3366,584,2014,2396,1902,4569,3002,1938,3998,4093,1899,3071,2815,1974,302,1641,2836,565,264,1332,3319,3689,2181,3873,4883,3849,1991,4633,4556,3866,142,2903,3181,740,3311,2071,280,714,2440,3950,290,3580,738,1604,3631,1989,1299,836,1913,224,1066,1741,1551,1735,4601,2024,4570,4192,1723,3949,3696,1419,1760,697,4764,3405,4443,199,717,4568,3252,2016,2151,1741,2613,2736,4053,814,4282,3392,615,1998,3294,3663,559,4278,4626,55,1418,2056,3191,3181,1732,1887,2517,3180,2154,2166,3096,3930,2721,4332,427,4332,4237,3928,2262,4657,2202,922,3711,1921,4728,2236,2441,622,233,293,1466,1891,1222,3693,3261,2605,3486,102,3612,1897,2698,3524,3567,613,3834,1583,1482,4734,2339,752,1428,4121,3267,3518,4652,3119,1818,4596,3181,3159,4069,3375,3762,1386,3054,3052,67,2246,1493,2738,2835,4906,303,1107,3111,1525,1739,437,2941,545,1458,993,1871,640,4047,2017,4971,4917,701,4811,4335,3221,4187,4414,756,3069,3052,812,3135,928,1264,3356,4518,2136,2691,2638,3156,4909,2944,3920,4609,1856,654,4643,2932,309,3613,4479,4173,1848,165,1171,592,3233,3151,4009,3952,2624,38,2616,2056,841,1764,4667,1526,125,3963,933,3951,2151,2110,4666,1000,1985,3868,2735,635,277,1129,572,2136,980,2731,556,3012,2900,2180,1912,2799,1771,4441,2666,3958,4381,3677,4218,1276,3512,4868,4579,2307,3952,3544,651,1300,218,489,2837,3737,509,3421,879,4353,4695]
Fee: 655
考虑到以下因素，我正在尝试添加备注： -用户在第M天购买，并在第Index天出售。
尽管我确实误解了这个问题上的州，但改道了。请提出更好的建议。
谢谢。
答案 0 :(得分：3)
您的解决方案具有二次复杂度，因此会产生TLE。最佳解决方案具有线性复杂度。这是从二次解中得出线性的步骤。
二次时间解决方案
d[i]-您最多可以获得第i个头寸（价格）的最大利润。
d[i] = max(d[i-1], max_{0<=j<i-1}(d[j] - price[j+1]) + price[i] - fee)
在位置i上我们什么也不能做，这意味着位置i上的最大利润是之前的最大利润，因此{{1}中的第一个术语d[i-1] }。
在位置max，我们可以出售股票。这样我们得到i。但是，我们必须事先在任何小于+price[i] - fee的头寸j+1处购买股票。通过购买此股票，我们将支付i，因此我们的利润将被price[j]改变。当我们在头寸-price[j+1]买入股票时，我们可以获取最大利润，直至头寸j+1（期限j），然后将其与上次交易d[j]的利润相加。最后，我们不知道购买股票的最佳时间-price[j+1] + price[i] - fee是什么，因此我们检查每个小于当前头寸j+1的头寸j，因此检查第二个{{1} }
线性时间解决方案
破坏我们复杂性的是每个i的{{1}}计算。因此，我们可以尝试使用动态编程来更新此最大值。让我们定义
max_{0<=j<i-1}
很显然，我们可以根据先前的值来计算max_{0<=j<i-1}(d[j] - price[j+1])，所以
i
，我们对h[i] = max_{0<=j<i}(d[j] - price[j+1]),的原始公式变为：
h[i]。现在，对于每个h[i] = max(h[i-1], d[i-1] - price[i])，我们只需更新d和d[i] = max(d[i-1], h[i-1] + price[i] - fee)，这样我们就有了线性时间复杂度，其中的解决方案是i，考虑了d的价格。
以下是完整的代码：
h 时间复杂度为d[n]，空间复杂度为n。
请注意，我们仅使用先前的值public int MaxProfit(int[] prices, int fee) {
int n = prices.Length;
if (n == 0) return 0;
int[] d = new int[n]; int[] h = new int[n];
d[0] = 0; h[0] = -prices[0];
for(int i = 1; i < n; i++){
h[i] = Math.Max(h[i-1], d[i-1] - prices[i]);
d[i] = Math.Max(d[i-1], h[i-1] + prices[i] - fee);
}
return d[n-1];
}
和O(n)，因此可以将空间复杂度进一步降低到O(n)。
d[i-1] 时间复杂度为h[i-1]，空间复杂度为O(1)。
答案 1 :(得分：2)
状态机-解决诸如 1 这样的库存问题时的另一种思维方式。
假设我们有一个状态机：
|-----|
| end |
|-----|
^
|
|
| buy
|--- |------| ----------------> |------| ---|
rest | | s0 | | s1 | | rest
|--> |------| <---------------- |------| <--|
sell
共有3个选项：购买，出售或休息。
s0州时，我们可以休息或购买股票； s1州时，我们可以休息或出售股票； 在每个步骤中，我们都想采取使利润最大化的行动。因此，我们可以将状态转换表示如下：
s0 = Math.max(s0, s1 + price[i]); // rest or sell from s1
s1 = Math.max(s1, s0 - price[i] - fee); // rest or buy with a fee from s0
最终解决方案：
public int maxProfit(int[] prices, int fee) {
int s0 = 0;
int s1 = Integer.MIN_VALUE;
for (int p : prices) {
s0 = Math.max(s0, s1 + p);
s1 = Math.max(s1, s0 - p - fee);
}
return s0;
}
时间复杂度：O(n)。空间复杂度：O(1)。
1-最初的想法来自here。