370 lines
11 KiB
Markdown
370 lines
11 KiB
Markdown
## 题目地址
|
||
https://leetcode.com/problems/maximum-subarray/
|
||
|
||
## 题目描述
|
||
```
|
||
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.
|
||
```
|
||
|
||
## 思路
|
||
|
||
这道题求解连续最大子序列和，以下从时间复杂度角度分析不同的解题思路。
|
||
|
||
#### 解法一 - 暴力解 （暴力出奇迹， 噢耶！）
|
||
一般情况下，先从暴力解分析，然后再进行一步步的优化。
|
||
|
||
**原始暴力解：**（超时）
|
||
|
||
求子序列和，那么我们要知道子序列的首尾位置，然后计算首尾之间的序列和。用2个for循环可以枚举所有子序列的首尾位置。
|
||
然后用一个for循环求解序列和。这里时间复杂度太高，`O(n^3)`.
|
||
|
||
#### 复杂度分析
|
||
- *时间复杂度：* `O(n^3) - n 是数组长度`
|
||
- *空间复杂度：* `O(1)`
|
||
|
||
#### 解法二 - 前缀和 + 暴力解
|
||
**优化暴力解：** (震惊，居然AC了）
|
||
|
||
在暴力解的基础上，用前缀和我们可以优化到暴力解`O(n^2)`, 这里以空间换时间。
|
||
这里可以使用原数组表示`prefixSum`, 省空间。
|
||
|
||
求序列和可以用前缀和（`prefixSum`) 来优化，给定子序列的首尾位置`（l, r),`
|
||
那么序列和 `subarraySum=prefixSum[r] - prefixSum[l - 1];`
|
||
用一个全局变量`maxSum`, 比较每次求解的子序列和，`maxSum = max(maxSum, subarraySum)`.
|
||
|
||
#### 复杂度分析
|
||
- *时间复杂度：* `O(n^2) - n 是数组长度`
|
||
- *空间复杂度：* `O(n) - prefixSum 数组空间为n`
|
||
|
||
>如果用更改原数组表示前缀和数组，空间复杂度降为`O(1)`
|
||
|
||
但是时间复杂度还是太高，还能不能更优化。答案是可以，前缀和还可以优化到`O(n)`.
|
||
|
||
#### 解法三 - 优化前缀和 - from [**@lucifer**](https://github.com/azl397985856)
|
||
|
||
我们定义函数` S(i)` ，它的功能是计算以 `0（包括 0）`开始加到 `i（包括 i）`的值。
|
||
|
||
那么 `S(j) - S(i - 1)` 就等于 从 `i` 开始（包括 i）加到 `j`（包括 j）的值。
|
||
|
||
我们进一步分析，实际上我们只需要遍历一次计算出所有的 `S(i)`, 其中 `i = 0,1,2....,n-1。`
|
||
然后我们再减去之前的` S(k)`,其中 `k = 0，1，i - 1`，中的最小值即可。 因此我们需要
|
||
用一个变量来维护这个最小值，还需要一个变量维护最大值。
|
||
|
||
#### 复杂度分析
|
||
- *时间复杂度：* `O(n) - n 是数组长度`
|
||
- *空间复杂度：* `O(1)`
|
||
|
||
#### 解法四 - [分治法](https://www.wikiwand.com/zh-hans/%E5%88%86%E6%B2%BB%E6%B3%95)
|
||
|
||
我们把数组`nums`以中间位置（`m`)分为左（`left`)右(`right`)两部分. 那么有，
|
||
`left = nums[0]...nums[m - 1]` 和 `right = nums[m + 1]...nums[n-1]`
|
||
|
||
最大子序列和的位置有以下三种情况：
|
||
1. 考虑中间元素`nums[m]`, 跨越左右两部分，这里从中间元素开始，往左求出后缀最大，往右求出前缀最大, 保持连续性。
|
||
2. 不考虑中间元素，最大子序列和出现在左半部分，递归求解左边部分最大子序列和
|
||
3. 不考虑中间元素，最大子序列和出现在右半部分，递归求解右边部分最大子序列和
|
||
|
||
分别求出三种情况下最大子序列和，三者中最大值即为最大子序列和。
|
||
|
||
举例说明，如下图：
|
||
|
||
|
||
#### 复杂度分析
|
||
- *时间复杂度：* `O(nlogn) - n 是数组长度`
|
||
- *空间复杂度：* `O(logn)` - 因为调用栈的深度最多是logn。
|
||
|
||
#### 解法五 - [动态规划](https://www.wikiwand.com/zh-hans/%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92)
|
||
动态规划的难点在于找到状态转移方程，
|
||
|
||
`dp[i] - 表示到当前位置 i 的最大子序列和`
|
||
|
||
状态转移方程为：
|
||
`dp[i] = max(dp[i - 1] + nums[i], nums[i])`
|
||
|
||
初始化：`dp[0] = nums[0]`
|
||
|
||
从状态转移方程中，我们只关注前一个状态的值，所以不需要开一个数组记录位置所有子序列和，只需要两个变量，
|
||
|
||
`currMaxSum - 累计最大和到当前位置i`
|
||
|
||
`maxSum - 全局最大子序列和`:
|
||
|
||
- `currMaxSum = max(currMaxSum + nums[i], nums[i])`
|
||
- `maxSum = max(currMaxSum, maxSum)`
|
||
|
||
如图：
|
||
|
||
|
||
#### 复杂度分析
|
||
- *时间复杂度:* `O(n) - n 是数组长度`
|
||
- *空间复杂度:* `O(1)`
|
||
|
||
## 关键点分析
|
||
1. 暴力解，列举所有组合子序列首尾位置的组合，求解最大的子序列和, 优化可以预先处理，得到前缀和
|
||
2. 分治法，每次从中间位置把数组分为左右中三部分， 分别求出左右中（这里中是包括中间元素的子序列）最大和。对左右分别深度递归，三者中最大值即为当前最大子序列和。
|
||
3. 动态规划，找到状态转移方程，求到当前位置最大和。
|
||
|
||
## 代码 (`Java/Python3/Javascript`)
|
||
#### 解法二 - 前缀和 + 暴力
|
||
*Java code*
|
||
```java
|
||
class MaximumSubarrayPrefixSum {
|
||
public int maxSubArray(int[] nums) {
|
||
int len = nums.length;
|
||
int maxSum = Integer.MIN_VALUE;
|
||
int sum = 0;
|
||
for (int i = 0; i < len; i++) {
|
||
sum = 0;
|
||
for (int j = i; j < len; j++) {
|
||
sum += nums[j];
|
||
maxSum = Math.max(maxSum, sum);
|
||
}
|
||
}
|
||
return maxSum;
|
||
}
|
||
}
|
||
```
|
||
*Python3 code* `(TLE)`
|
||
```python
|
||
import sys
|
||
class Solution:
|
||
def maxSubArray(self, nums: List[int]) -> int:
|
||
n = len(nums)
|
||
maxSum = -sys.maxsize
|
||
sum = 0
|
||
for i in range(n):
|
||
sum = 0
|
||
for j in range(i, n):
|
||
sum += nums[j]
|
||
maxSum = max(maxSum, sum)
|
||
|
||
return maxSum
|
||
```
|
||
|
||
*Javascript code* from [**@lucifer**](https://github.com/azl397985856)
|
||
|
||
```javascript
|
||
function LSS(list) {
|
||
const len = list.length;
|
||
let max = -Number.MAX_VALUE;
|
||
let sum = 0;
|
||
for (let i = 0; i < len; i++) {
|
||
sum = 0;
|
||
for (let j = i; j < len; j++) {
|
||
sum += list[j];
|
||
if (sum > max) {
|
||
max = sum;
|
||
}
|
||
}
|
||
}
|
||
|
||
return max;
|
||
}
|
||
```
|
||
#### 解法三 - 优化前缀和
|
||
*Java code*
|
||
```java
|
||
class MaxSumSubarray {
|
||
public int maxSubArray3(int[] nums) {
|
||
int maxSum = nums[0];
|
||
int sum = 0;
|
||
int minSum = 0;
|
||
for (int num : nums) {
|
||
// prefix Sum
|
||
sum += num;
|
||
// update maxSum
|
||
maxSum = Math.max(maxSum, sum - minSum);
|
||
// update minSum
|
||
minSum = Math.min(minSum, sum);
|
||
}
|
||
return maxSum;
|
||
}
|
||
}
|
||
```
|
||
*Python3 code*
|
||
```python
|
||
class Solution:
|
||
def maxSubArray(self, nums: List[int]) -> int:
|
||
n = len(nums)
|
||
maxSum = nums[0]
|
||
minSum = sum = 0
|
||
for i in range(n):
|
||
sum += nums[i]
|
||
maxSum = max(maxSum, sum - minSum)
|
||
minSum = min(minSum, sum)
|
||
|
||
return maxSum
|
||
```
|
||
|
||
*Javascript code* from [**@lucifer**](https://github.com/azl397985856)
|
||
```javascript
|
||
function LSS(list) {
|
||
const len = list.length;
|
||
let max = list[0];
|
||
let min = 0;
|
||
let sum = 0;
|
||
for (let i = 0; i < len; i++) {
|
||
sum += list[i];
|
||
if (sum - min > max) max = sum - min;
|
||
if (sum < min) {
|
||
min = sum;
|
||
}
|
||
}
|
||
|
||
return max;
|
||
}
|
||
```
|
||
|
||
#### 解法四 - 分治法
|
||
|
||
*Java code*
|
||
```java
|
||
class MaximumSubarrayDivideConquer {
|
||
public int maxSubArrayDividConquer(int[] nums) {
|
||
if (nums == null || nums.length == 0) return 0;
|
||
return helper(nums, 0, nums.length - 1);
|
||
}
|
||
private int helper(int[] nums, int l, int r) {
|
||
if (l > r) return Integer.MIN_VALUE;
|
||
int mid = (l + r) >>> 1;
|
||
int left = helper(nums, l, mid - 1);
|
||
int right = helper(nums, mid + 1, r);
|
||
int leftMaxSum = 0;
|
||
int sum = 0;
|
||
// left surfix maxSum start from index mid - 1 to l
|
||
for (int i = mid - 1; i >= l; i--) {
|
||
sum += nums[i];
|
||
leftMaxSum = Math.max(leftMaxSum, sum);
|
||
}
|
||
int rightMaxSum = 0;
|
||
sum = 0;
|
||
// right prefix maxSum start from index mid + 1 to r
|
||
for (int i = mid + 1; i <= r; i++) {
|
||
sum += nums[i];
|
||
rightMaxSum = Math.max(sum, rightMaxSum);
|
||
}
|
||
// max(left, right, crossSum)
|
||
return Math.max(leftMaxSum + rightMaxSum + nums[mid], Math.max(left, right));
|
||
}
|
||
}
|
||
```
|
||
|
||
*Python3 code*
|
||
|
||
```python
|
||
import sys
|
||
class Solution:
|
||
def maxSubArray(self, nums: List[int]) -> int:
|
||
return self.helper(nums, 0, len(nums) - 1)
|
||
def helper(self, nums, l, r):
|
||
if l > r:
|
||
return -sys.maxsize
|
||
mid = (l + r) // 2
|
||
left = self.helper(nums, l, mid - 1)
|
||
right = self.helper(nums, mid + 1, r)
|
||
left_suffix_max_sum = right_prefix_max_sum = 0
|
||
sum = 0
|
||
for i in reversed(range(l, mid)):
|
||
sum += nums[i]
|
||
left_suffix_max_sum = max(left_suffix_max_sum, sum)
|
||
sum = 0
|
||
for i in range(mid + 1, r + 1):
|
||
sum += nums[i]
|
||
right_prefix_max_sum = max(right_prefix_max_sum, sum)
|
||
cross_max_sum = left_suffix_max_sum + right_prefix_max_sum + nums[mid]
|
||
return max(cross_max_sum, left, right)
|
||
```
|
||
|
||
*Javascript code* from [**@lucifer**](https://github.com/azl397985856)
|
||
|
||
```javascript
|
||
function helper(list, m, n) {
|
||
if (m === n) return list[m];
|
||
let sum = 0;
|
||
let lmax = -Number.MAX_VALUE;
|
||
let rmax = -Number.MAX_VALUE;
|
||
const mid = ((n - m) >> 1) + m;
|
||
const l = helper(list, m, mid);
|
||
const r = helper(list, mid + 1, n);
|
||
for (let i = mid; i >= m; i--) {
|
||
sum += list[i];
|
||
if (sum > lmax) lmax = sum;
|
||
}
|
||
|
||
sum = 0;
|
||
|
||
for (let i = mid + 1; i <= n; i++) {
|
||
sum += list[i];
|
||
if (sum > rmax) rmax = sum;
|
||
}
|
||
|
||
return Math.max(l, r, lmax + rmax);
|
||
}
|
||
|
||
function LSS(list) {
|
||
return helper(list, 0, list.length - 1);
|
||
}
|
||
```
|
||
|
||
#### 解法五 - 动态规划
|
||
|
||
*Java code*
|
||
```java
|
||
class MaximumSubarrayDP {
|
||
public int maxSubArray(int[] nums) {
|
||
int currMaxSum = nums[0];
|
||
int maxSum = nums[0];
|
||
for (int i = 1; i < nums.length; i++) {
|
||
currMaxSum = Math.max(currMaxSum + nums[i], nums[i]);
|
||
maxSum = Math.max(maxSum, currMaxSum);
|
||
}
|
||
return maxSum;
|
||
}
|
||
}
|
||
```
|
||
|
||
*Python3 code*
|
||
```python
|
||
class Solution:
|
||
def maxSubArray(self, nums: List[int]) -> int:
|
||
n = len(nums)
|
||
max_sum_ending_curr_index = max_sum = nums[0]
|
||
for i in range(1, n):
|
||
max_sum_ending_curr_index = max(max_sum_ending_curr_index + nums[i], nums[i])
|
||
max_sum = max(max_sum_ending_curr_index, max_sum)
|
||
|
||
return max_sum
|
||
```
|
||
|
||
*Javascript code* from [**@lucifer**](https://github.com/azl397985856)
|
||
|
||
```javascript
|
||
function LSS(list) {
|
||
const len = list.length;
|
||
let max = list[0];
|
||
for (let i = 1; i < len; i++) {
|
||
list[i] = Math.max(0, list[i - 1]) + list[i];
|
||
if (list[i] > max) max = list[i];
|
||
}
|
||
|
||
return max;
|
||
}
|
||
```
|
||
|
||
## 扩展
|
||
- 如果数组是二维数组，求最大子数组的和？
|
||
- 如果要求最大子序列的乘积？
|
||
|
||
## 相似题
|
||
- [Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/)
|
||
- [Longest Turbulent Subarray](https://leetcode.com/problems/longest-turbulent-subarray/)
|