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665. Non-decreasing Array只允許修改一位數的非遞減數組

you pan 修改 with 暴力 ive ret nbsp 優化

[抄題]:

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.

[暴力解法]:

時間分析:

空間分析:

[優化後]:

時間分析:

空間分析:

[奇葩輸出條件]:

[奇葩corner case]:

[思維問題]:

不知道怎麽改啊

[一句話思路]:

既然只允許修改一位,“前天”是否異常,決定了應該修改“昨天”還是“今天”

[輸入量]:空: 正常情況:特大:特小:程序裏處理到的特殊情況:異常情況(不合法不合理的輸入):

[畫圖]:

技術分享圖片

[一刷]:

  1. 必須要有“昨天”異常的前提才有後續的操作。所以要把“昨天”之後的全都括起來

[二刷]:

[三刷]:

[四刷]:

[五刷]:

[五分鐘肉眼debug的結果]:

[總結]:

頭回見:既然只允許修改一位,“前天”是否異常,決定了應該修改“昨天”還是“今天”

[復雜度]:Time complexity: O(n) Space complexity: O(1)

[英文數據結構或算法,為什麽不用別的數據結構或算法]:

[關鍵模板化代碼]:

[其他解法]:

[Follow Up]:

[LC給出的題目變變變]:

[代碼風格] :

技術分享圖片
class Solution {
    
public boolean checkPossibility(int[] nums) { //cc if (nums == null || nums.length == 0) { return false; } //ini int count = 0; //for loop for (int i = 1; i < nums.length && count <= 1; i++) { if (nums[i - 1] > nums[i]) {count++; if (i - 2 < 0 || nums[i - 2] < nums[i]) { nums[i - 1] = nums[i]; }else { nums[i] = nums[i - 1]; }} } //return count <= 1 return count <= 1; } }
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665. Non-decreasing Array只允許修改一位數的非遞減數組