Description:
Solution:
I set the right and left pointers to the end of the array.
m is the pivot(first set the middle of the array).
If m is more than l, then we can throw out the left portion. Now, the new l is the next value of m.
Else, we can throw out the right portion, so the new r is the previous value of m.
When we get the nums[l] is less than nums[r], we return the res.
'LeetCode ๐๏ธ > Array' ์นดํ ๊ณ ๋ฆฌ์ ๋ค๋ฅธ ๊ธ
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