[LeetCode] 665. Non-decreasing Array

轉自LeetCode

Given an array with n integers, your task is to check if it could become non-decreasing by modifying at most 1 element.

We define an array is non-decreasing if array[i] <= array[i + 1] holds for every i (1 <= i < n).

Example 1:

Input: [4,2,3]
Output: True
Explanation: You could modify the first 4 to 1 to get a non-decreasing array.

Example 2:

Input: [4,2,1]
Output: False
Explanation: You can't get a non-decreasing array by modify at most one element.

Note: The n belongs to [1, 10,000].

<Solution>

想法如下

• 找到數字變小的地方，不難，就是按順序歷遍找就好。假設 nums[i-1] > nums[i]

• 比較難的地方是，要怎麼修正。如果數列長度小於 2，那直接修正其一即可

• 數列長度大於等於2，然後 nums[i-2] 小於等於 nums[i]，那就只要把 nums[i-1] 改成 nums[i] 就好

• 數列長度大於等於2，然後 nums[i-2] 大於 nums[i]，那麼就要把 nums[i] 改成 nums[i-1]

• 改完後繼續歷遍，最後的 cnt 如果大於 1，就回傳 false，反之回傳 true

code 如下

Java

	class Solution {
	public boolean checkPossibility(int[] nums) {
	int cnt = 0;
	for(int i = 1; i < nums.length; i++) {
	if(nums[i - 1] > nums[i]) {
	++cnt;
	if(i-2 < 0 || nums[i-2] <= nums[i]) {
	nums[i-1] = nums[i];
	}
	else {
	nums[i] = nums[i-1];
	}
	}
	}
	return cnt <= 1;
	}
	}

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