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JavaArithmetic/src/LeetCode/LeetCode209.java at master · feinibuai/JavaArithmetic · GitHub
feinibuai
JavaArithmetic
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LeetCode
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LeetCode209.java
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JavaArithmetic
/
src
/
LeetCode
/
LeetCode209.java
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package
LeetCode
;
public
class
LeetCode209
{
// https://leetcode.com/problems/minimum-size-subarray-sum/description/
// 暴力解法
// 该方法在 Leetcode 中会超时！
// 时间复杂度: O(n^3)
// 空间复杂度: O(1)
public
int
minSubArrayLen
(
int
s
,
int
[]
nums
) {
if
(
s
<=
0
||
nums
==
null
)
throw
new
IllegalArgumentException
(
"Illigal Arguments"
);
int
res
=
nums
.
length
+
1
;
for
(
int
l
=
0
;
l
<
nums
.
length
;
l
++)
for
(
int
r
=
l
;
r
<
nums
.
length
;
r
++) {
int
sum
=
0
;
for
(
int
i
=
l
;
i
<=
r
;
i
++)
sum
+=
nums
[
i
];
if
(
sum
>=
s
)
res
=
Math
.
min
(
res
,
r
-
l
+
1
);
}
// 无解的情况
if
(
res
==
nums
.
length
+
1
)
return
0
;
return
res
;
}
// 优化暴力解
// 时间复杂度: O(n^2)
// 空间复杂度: O(n)
public
int
minSubArrayLen2
(
int
s
,
int
[]
nums
) {
if
(
s
<=
0
||
nums
==
null
)
throw
new
IllegalArgumentException
(
"Illigal Arguments"
);
// sums[i]存放nums[0...i-1]的和
int
[]
sums
=
new
int
[
nums
.
length
+
1
];
sums
[
0
] =
0
;
for
(
int
i
=
1
;
i
<=
nums
.
length
;
i
++)
sums
[
i
] =
sums
[
i
-
1
] +
nums
[
i
-
1
];
int
res
=
nums
.
length
+
1
;
for
(
int
l
=
0
;
l
<
nums
.
length
;
l
++)
for
(
int
r
=
l
;
r
<
nums
.
length
;
r
++){
// 使用sums[r+1] - sums[l] 快速获得nums[l...r]的和
if
(
sums
[
r
+
1
] -
sums
[
l
] >=
s
)
res
=
Math
.
min
(
res
,
r
-
l
+
1
);
}
// 无解的情况
if
(
res
==
nums
.
length
+
1
)
return
0
;
return
res
;
}
// 滑动窗口的思路
// 时间复杂度: O(n)
// 空间复杂度: O(1)
public
int
minSubArrayLen3
(
int
s
,
int
[]
nums
) {
if
(
s
<=
0
||
nums
==
null
)
throw
new
IllegalArgumentException
(
"Illigal Arguments"
);
int
l
=
0
,
r
= -
1
;
// nums[l...r]为我们的滑动窗口
int
sum
=
0
;
int
res
=
nums
.
length
+
1
;
while
(
l
<
nums
.
length
){
// 窗口的左边界在数组范围内,则循环继续
if
(
r
+
1
<
nums
.
length
&&
sum
<
s
)
sum
+=
nums
[++
r
];
else
// r已经到头 或者 sum >= s
sum
-=
nums
[
l
++];
if
(
sum
>=
s
)
res
=
Math
.
min
(
res
,
r
-
l
+
1
);
}
// 无解的情况
if
(
res
==
nums
.
length
+
1
)
return
0
;
return
res
;
}
// 另外一个滑动窗口的实现, 仅供参考
// 时间复杂度: O(n)
// 空间复杂度: O(1)
public
int
minSubArrayLen4
(
int
s
,
int
[]
nums
) {
if
(
s
<=
0
||
nums
==
null
)
throw
new
IllegalArgumentException
(
"Illigal Arguments"
);
int
l
=
0
,
r
= -
1
;
// [l...r]为我们的窗口
int
sum
=
0
;
int
res
=
nums
.
length
+
1
;
while
(
r
+
1
<
nums
.
length
){
// 窗口的右边界无法继续扩展了, 则循环继续
while
(
r
+
1
<
nums
.
length
&&
sum
<
s
)
sum
+=
nums
[++
r
];
if
(
sum
>=
s
)
res
=
Math
.
min
(
res
,
r
-
l
+
1
);
while
(
l
<
nums
.
length
&&
sum
>=
s
){
sum
-=
nums
[
l
++];
if
(
sum
>=
s
)
res
=
Math
.
min
(
res
,
r
-
l
+
1
);
}
}
if
(
res
==
nums
.
length
+
1
)
return
0
;
return
res
;
}
// 时间复杂度: O(nlogn)
// 空间复杂度: O(n)
// 二分搜索
public
int
minSubArrayLen5
(
int
s
,
int
[]
nums
) {
if
(
s
<=
0
||
nums
==
null
)
throw
new
IllegalArgumentException
(
"Illigal Arguments"
);
// sums[i]存放nums[0...i-1]的和
int
[]
sums
=
new
int
[
nums
.
length
+
1
];
sums
[
0
] =
0
;
for
(
int
i
=
1
;
i
<=
nums
.
length
;
i
++)
sums
[
i
] =
sums
[
i
-
1
] +
nums
[
i
-
1
];
int
res
=
nums
.
length
+
1
;
for
(
int
l
=
0
;
l
<
nums
.
length
-
1
;
l
++){
// Java类库中没有内置的lowerBound方法，
// 我们需要自己实现一个基于二分搜索的lowerBound:)
int
r
=
lowerBound
(
sums
,
sums
[
l
] +
s
);
if
(
r
!=
sums
.
length
){
res
=
Math
.
min
(
res
,
r
-
l
);
}
}
if
(
res
==
nums
.
length
+
1
)
return
0
;
return
res
;
}
// 在有序数组nums中寻找大于等于target的最小值
// 如果没有（nums数组中所有值都小于target），则返回nums.length
private
int
lowerBound
(
int
[]
nums
,
int
target
){
if
(
nums
==
null
/*|| !isSorted(nums)*/
)
throw
new
IllegalArgumentException
(
"Illegal argument nums in lowerBound."
);
int
l
=
0
,
r
=
nums
.
length
;
// 在nums[l...r)的范围里寻找解
while
(
l
!=
r
){
int
mid
=
l
+ (
r
-
l
) /
2
;
if
(
nums
[
mid
] >=
target
)
r
=
mid
;
else
l
=
mid
+
1
;
}
return
l
;
}
public
static
void
main
(
String
[]
args
) {
int
[]
nums
= {
2
,
3
,
1
,
2
,
4
,
3
};
int
s
=
7
;
System
.
out
.
println
((
new
LeetCode209
()).
minSubArrayLen
(
s
,
nums
));
}
}
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