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Twitter OA

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rwei 发表于 2014-12-23 06:27:32 | 显示全部楼层 |阅读模式

2014(10-12月) 码农类 本科 全职@Twitter - 猎头 - 在线笔试 |Pass

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Twitter OA:. 鍥磋鎴戜滑@1point 3 acres

(1)

Twosum from Leetcode.鏈枃鍘熷垱鑷1point3acres璁哄潧
. 1point3acres.com/bbs
第一題上 Leetcode 看. Waral 鍗氬鏈夋洿澶氭枃绔,

https://oj.leetcode.com/problems/two-sum/. 鍥磋鎴戜滑@1point 3 acres
.鏈枃鍘熷垱鑷1point3acres璁哄潧
(2)
. 涓浜-涓夊垎-鍦帮紝鐙鍙戝竷
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? . 1point 3acres 璁哄潧

'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 .鐣欏璁哄潧-涓浜-涓夊垎鍦
An integer N
Next line contains the N bits, separated by spaces: d[0] d[1] ... d[N - 1]

Output:
S . 鍥磋鎴戜滑@1point 3 acres

Constraints:
1 <= N <= 100000
d can only be 0 or 1f
0 <= L <= R < n .1point3acres缃
. 涓浜-涓夊垎-鍦帮紝鐙鍙戝竷
Sample Input:
8
1 0 0 1 0 0 1 0

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


這題可以這麼做:

public static void flipBits(int[] a) {

        int maxDiff = 0;
        int flipStartIndex = 0;
        int flipEndIndex = 0;
        int onesToFlip = 0;
        int totalNumberOfOnes = 0;

        int currentDiff = 0;
        int currentStart = 0;
        int currentOnesToFlip = 0;

        for (int i = 0; i < a.length; i++) {
            if (a == 0) {
                        currentDiff -= 1;
            } else {
                        currentDiff += 1;
                        currentOnesToFlip++;
                        totalNumberOfOnes++;
            }
            if (currentDiff < maxDiff) {
                        maxDiff = currentDiff;
                        flipStartIndex = currentStart;
                        flipEndIndex = i;
                        onesToFlip = currentOnesToFlip;
            } else if (currentDiff > 0) {
                        currentDiff = 0;
                        currentStart = i + 1;
                        currentOnesToFlip = 0;
            }
        }
        System.out.println(flipEndIndex - flipStartIndex + 1 - onesToFlip +   totalNumberOfOnes - onesToFlip);
}

Space O(1), Time O(n)


.鐣欏璁哄潧-涓浜-涓夊垎鍦
yannan 发表于 2015-2-7 11:37:31 | 显示全部楼层
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Warald_一亩三分地
我今天OA第二题和你一样!
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 楼主| rwei 发表于 2015-2-7 11:39:49 来自手机 | 显示全部楼层
关注一亩三分地微博:
Warald
yannan 发表于 2015-2-7 11:37
我今天OA第二题和你一样!
. Waral 鍗氬鏈夋洿澶氭枃绔,
估計最近的都是吧…命中了求加分
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