7.1 MaxProfit

求股票最大收益。算法导论上的问题
A zero-indexed array A consisting of N integers is given. It contains daily prices of a stock share for a period of N consecutive days. If a single share was bought on day P and sold on day Q, where 0 ≤ P ≤ Q < N, then the profit of such transaction is equal to A[Q] − A[P], provided that A[Q] ≥ A[P]. Otherwise, the transaction brings loss of A[P] − A[Q].
For example, consider the following array A consisting of six elements such that:
A[0] = 23171
A[1] = 21011
A[2] = 21123
A[3] = 21366
A[4] = 21013
A[5] = 21367
If a share was bought on day 0 and sold on day 2, a loss of 2048 would occur because A[2] − A[0] = 21123 − 23171 = −2048. If a share was bought on day 4 and sold on day 5, a profit of 354 would occur because A[5] − A[4] = 21367 − 21013 = 354. Maximum possible profit was 356. It would occur if a share was bought on day 1 and sold on day 5.
Write a function,
class Solution { public int solution(int[] A); }
that, given a zero-indexed array A consisting of N integers containing daily prices of a stock share for a period of N consecutive days, returns the maximum possible profit from one transaction during this period. The function should return 0 if it was impossible to gain any profit.
For example, given array A consisting of six elements such that:
A[0] = 23171
A[1] = 21011
A[2] = 21123
A[3] = 21366
A[4] = 21013
A[5] = 21367
the function should return 356, as explained above.
Assume that:
N is an integer within the range [0..400,000];
each element of array A is an integer within the range [0..200,000].
Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

Solution

扫描算法是扫描数据，记录下分段冠军依次PK，最后获胜者即为总冠军。难点在于找到分段的条件。本例的分段条件是A[i] < minCur，当A[i]比目前最小值更小的时候，有可能出现之后的数据-A[i]比当前的值大。
另一个经典问题连续最大值，分段条件是当前的sum=0时要开始新的分段了。

class Solution {    public int solution(int[] A) {        if(A.length < 2) return 0;        int maxEarn = 0;        int minCur = A[0];        for(int i=1; i<A.length; i++){            if(A[i] < minCur){                minCur = A[i];            }            if(A[i]-minCur > maxEarn){                maxEarn = A[i] - minCur;            }        }        return maxEarn;    }}