求股票最大收益。算法導論上的問題
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

掃描算法是掃描數據，記錄下分段冠軍依次PK，最後獲勝者即爲總冠軍。難點在於找到分段的條件。本例的分段條件是A[i] < minCur，當A[i]比目前最小值更小的時候，有可能出現之後的數據-A[i]比當前的值大。
另一個經典問題連續最大值，分段條件是當前的sum=0時要開始新的分段了。

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;
    }
}