为什么80%的码农都做不了架构师？>>> 
问题：
Say you have an array for which the ith element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete at most k transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
解决：
【题意】用一个数组表示股票每天的价格，数组的第i个数表示股票在第i天的价格。最多交易k次，手上最多只能持有一支股票，求最大收益。
① 特殊动态规划法。我们用一个局部最优解和全局最优解表示到第i天进行j次的收益，这就是该动态规划的特殊之处。
用local[i][j]表示到达第i天时，最多进行j次交易的局部最优解；用global[i][j]表示到达第i天时，最多进行j次的全局最优解。它们二者的关系如下（其中diff = prices[i] – prices[i – 1]）：
local[i][j] = max(global[i – 1][j – 1] + max(diff, 0), local[i – 1][j] + diff)
global[i][j] = max(global[i – 1][j], local[i][j])
其中的local[i – 1][j] + diff就是为了避免第i天交易和第i-1天交易合并成一次交易而少一次交易收益。时间O(n)，空间O(k)。
class Solution { //9ms
public int maxProfit(int k, int[] prices) {
if (prices.length < 2) return 0;
if (k >= prices.length) return maxProfit(prices);//k大于天数时，就退化为II，利用贪心算法查找最大利益,k=2时，就是III的情况，使用双向动态规划
int[] local = new int[k + 1];
int[] global = new int[k + 1];
for (int i = 1;i < prices.length;i ++){
int diff = prices[i] - prices[i - 1];
for (int j = k;j > 0;j --){
local[j] = Math.max(global[j - 1] + Math.max(diff,0),local[j] + diff);
global[j] = Math.max(global[j],local[j]);
}
}
return global[k];
}
public int maxProfit(int[] prices){
int max = 0;
for (int i = 1;i < prices.length;i ++){
if (prices[i] > prices[i - 1]){
max += prices[i] - prices[i - 1];
}
}
return max;
}
}
② 在discuss中看到的。。。
class Solution { //3ms
public int maxProfit(int k, int[] prices) {
if (prices.length < 2 || k <= 0){
return 0;
}
if (k == 1000000000){//如果不加，就不能通过OJ
return 1648961;
}
int[] local = new int[k + 1];
int[] global = new int[k + 1];
for (int i = 0;i < prices.length - 1;i ++){
int diff = prices[i + 1] - prices[i];
for (int j = k;j >= 1;j --){
local[j] = Math.max(global[j - 1] + Math.max(diff,0),local[j] + diff);
global[j] = Math.max(local[j],global[j]);
}
}
return global[k];
}
}