[LeetCode] 1004. Max Consecutive Ones III

 轉自LeetCode

Given a binary array nums and an integer k, return the maximum number of consecutive 1's in the array if you can flip at most k 0's.

 

Example 1:

Input: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2
Output: 6
Explanation: [1,1,1,0,0,1,1,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

Example 2:

Input: nums = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], k = 3
Output: 10
Explanation: [0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.

 

Constraints:

• 1 <= nums.length <= 105
• nums[i] is either 0 or 1.
• 0 <= k <= nums.length

Solution

485. Max Consecutive Ones 的衍生題

之前也有在 FB 的面試問到這題，可以用 sliding window 來解

關鍵的地方是要用個 counter 記錄目前已經用了幾個 0 了

如果 counter 小於零，那就要移動 left index，直到 counter 大於等於 0

kotlin

	class Solution {
	fun longestOnes(nums: IntArray, k: Int): Int {
	var left = -1
	var ans = 0
	val map = mutableMapOf<Int,Int>()
	var count = k
	for(right in nums.indices) {
	if(nums[right] == 0) {
	--count
	}
	while(count < 0) {
	++left
	if(nums[left] == 0) {
	++count
	}
	}
	ans = Math.max(ans, right-left)
	}
	return ans
	}
	}

view raw max_consecutive_ones_III.kt hosted with ❤ by GitHub