您的位置：首页 > 其它

Max Consecutive Ones

2017-09-19 11:48 281 查看

Q：

Input: [1,0,1,1,0]
Output: 4
Explanation: Flip the first zero will get the the maximum number of consecutive 1s. After flipping, the maximum number of consecutive 1s is 4.

原文链接：https://discuss.leetcode.com/topic/75445/java-clean-solution-easily-extensible-to-flipping-k-zero-and-follow-up-handled/2

The idea is to keep a window

[l, h]

that contains at most

zero

The following solution does not handle follow-up, because

nums[l]

will need to access
previous input stream

Time: O(n) Space: O(1)

public int findMaxConsecutiveOnes(int[] nums) {
int max = 0, zero = 0, k = 1; // flip at most k zero
for (int l = 0, h = 0; h < nums.length; h++) {
if (nums[h] == 0)
zero++;
while (zero > k)
if (nums[l++] == 0)
zero--;
max = Math.max(max, h - l + 1);
}
return max;
}

Now let's deal with follow-up, we need to store up to k

 indexes of zero within the window [l, h]
 so that we know where to move l
next when the window contains more than k
 zero. If the input stream is infinite, then the output could be extremely large because there could be super long consecutive ones. In that case we can use BigInteger
 for all indexes. For simplicity, here we will use int

Time: O(n) Space: O(k)

public int findMaxConsecutiveOnes(int[] nums) { [/code]

[/code]

int max = 0, k = 1; // flip at most k zero
Queue<Integer> zeroIndex = new LinkedList<>();
for (int l = 0, h = 0; h < nums.length; h++) {
if (nums[h] == 0)
zeroIndex.offer(h);
if (zeroIndex.size() > k)
l = zeroIndex.poll() + 1;
max = Math.max(max, h - l + 1);
}
return max;
}

Note that setting

k = 0

will give a solution to the earlier version Max Consecutive OnesFor

k = 1

we can apply the same idea to simplify the solution. Here

stores the index of zero within the window

[l, h]

so its role is similar to

Queue

in the above solution

public int findMaxConsecutiveOnes(int[] nums) {
int max = 0, q = -1;
for (int l = 0, h = 0; h < nums.length; h++) {
if (nums[h] == 0) {
l = q + 1;
q = h;
}
max = Math.max(max, h - l + 1);
}
return max;
}

内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理

标签：

相关文章推荐

新的分享

章节导航