【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
153. Find Minimum in Rotated Sorted Array
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e.,
[0,1,2,4,5,6,7]
might become
[4,5,6,7,0,1,2]
).
Find the minimum element.
You may assume no duplicate exists in the array.
Example 1:
Input:
[3,4,5,1,2]
Output:
1
Example 2:
Input:
[4,5,6,7,0,1,2]
Output:
0
/*
从头到尾遍历一遍为O(n)
*/
/*
问题：找旋转有序（增序）数组（不含重复数）中的最小数
分析：
旋转后，会形成两个有序左右子数组，且左子数组的数一定比右子数组的数大
如[3,4,5,1,2]
方法：二分查找
O(logn), O(1)
*/
class
Solution
{
public
:
int
findMin
(
vector
<
int
>
&
nums
)
{
int
left
=
0
,
right
=
nums
.
size
()
-
1
;
if
(
nums
[
left
]
>
nums
[
right
])
//若有旋转
{
while
(
left <
right
-
1
)
{
int
mid
=
(
left
+
right
)
/
2
;
if
(
nums
[
left
]
<
nums
[
mid
])
//若中间的数大，移动left指针到中间
left
=
mid
;
else
//若中间的数小，移动right指针到中间
right
=
mid
;
}
//退出时，left = right -1,left指向左子数组的末尾，right指向右子数组的开头
return
nums
[
right
];
}
else
//若无旋转，直接返回第一个数
return
nums
[
0
];
}
};
/*
其他写法
if (nums[mid] < nums[right]) //这种写法表示，右半段是有序的
right = mid;
else
left = mid;
*/
154
.
Find Minimum in Rotated Sorted Array II
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e.,
[0,1,2,4,5,6,7]
might become
[4,5,6,7,0,1,2]
).
Find the minimum element.
The array may contain duplicates.
Example 1:
Input:
[1,3,5]
Output:
1
Example 2:
Input:
[2,2,2,0,1]
Output:
0
Note:
-
Would allow duplicates affect the run-time complexity? How and why?
/*
问题：找旋转有序（增序）数组（含重复数）中的最小数
分析：
旋转后，会形成两个有序左右子数组，且左子数组的数一定比右子数组的数大(或等于)
如[2,2,2,0,1,1,2]
方法：二分查找，遇到相同数字时，left右移一位
平均O(logn),最坏O(n)
*/
class
Solution
{
public
:
int
findMin
(
vector
<
int
>
&
nums
)
{
if
(
nums
.
empty
())
return
0
;
if
(
nums
.
size
()
==
1
)
return
nums
[
0
];
int
res
=
nums
[
0
];
int
left
=
0
,
right
=
nums
.
size
()
-
1
;
if
(
nums
[
left
]
>=
nums
[
right
]) //若有旋转，由于有重复数字，故这里变为大于等于
{
while
(
left
<
right
-
1
)
{
int
mid
=
(
left
+
right
)
/
2
;
if
(
nums
[
left
]
<
nums
[
mid
])
//若中间的数大，移动left指针到中间
{
res
=
min
(
res
,
nums
[
left
]);
left
=
mid
;
}
else
if
(
nums
[
left
]
>
nums
[
mid
])
//若中间的数小，移动right指针到中间
{
res
=
min
(
res
,
nums
[
right
]);
right
=
mid
;
}
else
//若left与mid指向数字相等，将left右移一位,略过相同数字
left
++;
}
//退出时，left = right -1,left指向左子数组的末尾或右子数组的开头
res
=
min
(
res
,
nums
[
left
]);
res
=
min
(
res
,
nums
[
right
]);
return
res
;
}
else //若无旋转
return
nums
[
0
];
}
};