【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
53. Maximum Subarray
Given an integer array
nums
, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
Example:
Input:
[-2,1,-3,4,-1,2,1,-5,4],
Output:
6
Explanation:
[4,-1,2,1] has the largest sum = 6.
Follow up:
If you have figured out the O(n
) solution
, try coding another solution
using the divide and conquer approach,
which is more subtle.
//
最大子数组问题
/*
方法：动态规划
f(i) = a[i], i <0或f(i-1) <= 0;
f(i) = f(i-1) + a[i], i!=0, f(i-1) > 0
max(f[i]) 不太好想
*/
/*
方法一：动态规划
sum代表了包含nums[i]时（以nums[i]结尾的子数组）的最大和
不断更新sum和res sum = max(sum + num, num)
分析：时间复杂度
O
（
n
）
(
线性
)
*/
class
Solution
{
public
:
int
maxSubArray
(
vector
<
int
>&
nums
)
{
int
res
=
INT_MIN
,
sum
=
0
;
for
(
int
num
:
nums
)
{
sum = max(sum + num, num);
res = max(res, sum);
}
return
res
;
}
};
//
方法二：分治法
//
具体过程：将数组分为左子数组、右子数组、跨越中点的子数组问题
//
分析：时间复杂度
O(nlgn)
//参考资料：《算法导论》
class
Solution
{
public
:
int
maxSubArray
(
vector
<
int
>&
a
)
{
return
findMaxSubArray
(
a
,
0
,
a
.
size
()-
1
);
//
递归入口
}
//
递归函数：找
a[left...right]
的最大子数组（归并排序和快速排序中也用到了分治法，可以类比一下）
int
findMaxSubArray
(
vector
<
int
>&
a
,
int
left
,
int
right
)
//
递归函数
{
if
(
right
==
left
)
return
a
[
left
];
int
mid
=
(
left
+
right
)/
2
;
int
left_sum
=
findMaxSubArray
(
a
,
left
,
mid
);
int
right_sum
=
findMaxSubArray
(
a
,
mid
+
1
,
right
);
int
cross_sum
=
findMaxCrossingSubArray
(
a
,
left
,
mid
,
right
);
return
max
(
max
(
left_sum
,
right_sum
),
cross_sum
);
}
//
找跨中点的最大子数组，我们只需找出形如
A[i.. mid]
和
A[mid+ 1. .j]
的最大子数组，然后将其合并即可。
int
findMaxCrossingSubArray
(
vector
<
int
>&
a
,
int
left
,
int
mid
,
int
right
)
{
int
left_sum
,
sum
,
right_sum
;
sum
=
0
;
left_sum
=
a
[
mid
];
//
初始化为参与计算的第一个元素
for
(
int
i
=
mid
;
i
>=
left
;
i
--)
//从中间往左边遍历
{
sum
+=
a
[
i
];
if
(
sum
>
left_sum
)
left_sum
=
sum
;
}
sum
=
0
;
right_sum
=
a
[
mid
+
1
];
//
初始化
for
(
int
j
=
mid
+
1
;
j
<=
right
;
j
++)
//从中间往右边遍历
{
sum
+=
a
[
j
];
if
(
sum
>
right_sum
)
right_sum
=
sum
;
}
return
(
left_sum
+
right_sum
);
}
};