Twitter OA prepare: Flipping a bit

You are given a binary array with N elements: d[0], d[1], ... d[N - 1].
You can perform AT MOST one move on the array: choose any two integers [L, R], and flip all the elements between (and including) the L-th and R-th bits. L and R represent the left-most and right-most index of the bits marking the boundaries of the segment which you have decided to flip.

What is the maximum number of ‘1‘-bits (indicated by S) which you can obtain in the final bit-string? . more info on 1point3acres.com

‘Flipping‘ a bit means, that a 0 is transformed to a 1 and a 1 is transformed to a 0 (0->1,1->0).
Input Format
An integer N
Next line contains the N bits, separated by spaces: d[0] d[1] ... d[N - 1] 

Output:
S 

Constraints:
1 <= N <= 100000
d can only be 0 or 1f -google 1point3acres
0 <= L <= R < n
. 1point3acres.com/bbs
Sample Input:
8
1 0 0 1 0 0 1 0 . 1point3acres.com/bbs

Sample Output:
6 

Explanation: 

We can get a maximum of 6 ones in the given binary array by performing either of the following operations:
Flip [1, 5] ==> 1 1 1 0 1 1 1 0

分析：这道题无非就是在一个数组内，找一个区间，该区间  0的个数  与  1的个数  差值最大。如果我们把这个想成股票的话，0代表+1，1代表-1，那么这道题就转化成了Best Time to Buy and Sell Stock, 找0和1的个数差值最大就变成了找max profit。

因为需要找到这个区间，所以在Stock这道题的基础上还要做一定修改，记录区间边缘移动情况

 1 public int flipping(int[] A) {
 2    int local = 0;
 3    int global = 0;
 4    int localL = 0;
 5    int localR = 0;
 6    int globalL = 0;
 7    int globalR = 0;
 8    int OnesUnFlip = 0;   //those # of ones outside the chosen range
 9    for (int i=0; i<A.length; i++) {
10         int diff = 0;
11         if (A[i] == 0) diff = 1;
12         else diff = -1;
13
14         if (local + diff >= diff) {
15             local = local + diff;
16             localR = i;
17         }
18         else {
19             local = diff;
20             localL = i;
21             localR = i;
22         }
23
24         if (global < local) {
25             global = local;
26             globalL = localL;
27             globalR = localR;
28         }
29    }
30    for (int i=0; i<globalL; i++) {
31         if (A[i] == 1)
32             OnesUnflip ++;
33    }
34    for (int i=globalR+1; i<A.length; i++) {
35         if (A[i] == 1)
36             OnesUnflip ++;
37    }
38    return global + OnesUnflip;
39 }