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# 思路1 贪心

* 时间复杂度 $O(n)$
* 空间复杂度 $O(1)$

对于每个 i，需要让 nums\[i] <= nums\[i + 1]，如果有 nums\[i] > nums\[i + 1]，则要 modify

但是本题唯一的易错点就在这，

* 如果将 nums\[i] 缩小，可能会导致其无法融入前面已经遍历过的非递减子数列；

  &#x20; 3 4 2 3

  &#x20; 3 2 2 3
* 如果将 nums\[i + 1] 放大，可能会导致其后续的继续出现递减；

  &#x20; 3 4 2 3

  &#x20; 3 4 4 3

所以要采取贪心的策略，在遍历时，每次需要看连续的三个元素，也就是瞻前顾后，遵循以下两个原则：

* 需要尽可能不放大 nums\[i + 1]，这样会让后续非递减更困难；
* 如果缩小 nums\[i]，但不破坏前面的子序列的非递减性；

在遍历时，每次需要看连续的三个元素

遍历数组，如果遇到递减：

* 还能修改：
  * 修改方案1：将nums\[i]缩小至nums\[i + 1]；
  * 修改方案2：将nums\[i + 1]放大至nums\[i]；
* 不能修改了：直接返回false；

缩小时，必保证 nums\[i-1] <= nums\[i+1]。这样 nums\[i] 变小后，也比前一个大

否则，则将后面的放大


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