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yanglbmeyanglbme
committedJun 17, 2023
feat: add solutions to lc problem: No.0152
No.0152.Maximum Product Subarray
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‎solution/0100-0199/0152.Maximum Product Subarray/README.md

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@@ -43,16 +43,15 @@
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<!-- 这里可写通用的实现逻辑 -->
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考虑当前位置 i：
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**方法一：动态规划**
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- 如果是一个负数的话，那么我们希望以它前一个位置结尾的某个段的积也是个负数，这样可以负负得正，并且我们希望这个积尽可能「负得更多」，即尽可能小。
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- 如果是一个正数的话，我们更希望以它前一个位置结尾的某个段的积也是个正数，并且希望它尽可能地大。
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我们定义两个变量 $f$ 和 $g$，其中 $f$ 表示以 $nums[i]$ 结尾的乘积最大子数组的乘积，而 $g$ 表示以 $nums[i]$ 结尾的乘积最小子数组的乘积。初始时 $f$ 和 $g$ 都等于 $nums[0]$。答案为所有 $f$ 中的最大值。
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因此，分别维护 fmax 和 fmin。
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从 $i=1$ 开始，我们可以考虑将第 $i$ 个数 $nums[i]$ 添加到前面的乘积最大或者乘积最小的子数组乘积的后面，或者单独作为一个子数组乘积（即此时子数组长度只有 $1$）。我们将此前的乘积最大值记为 $ff$，乘积最小值记为 $gg$，那么 $f = \max(f, ff \times nums[i], gg \times nums[i])$，而 $g = \min(g, ff \times nums[i], gg \times nums[i])$。接下来，我们更新答案 $ans = \max(ans, f)$，然后继续计算下一个位置。
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- `fmax(i) = max(nums[i], fmax(i - 1) * nums[i], fmin(i - 1) * nums[i])`
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- `fmin(i) = min(nums[i], fmax(i - 1) * nums[i], fmin(i - 1) * nums[i])`
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- `res = max(fmax(i)), i∈[0, n)`
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最后的答案即为 $ans$。
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时间复杂度 $O(n)$，其中 $n$ 是数组 $nums$ 的长度。我们只需要遍历数组一次即可求得答案。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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@@ -63,13 +62,13 @@
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```python
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class Solution:
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def maxProduct(self, nums: List[int]) -> int:
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maxf = minf = res = nums[0]
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for num in nums[1:]:
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m, n = maxf, minf
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maxf = max(num, m * num, n * num)
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minf = min(num, m * num, n * num)
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res = max(res, maxf)
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return res
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ans = f = g = nums[0]
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for x in nums[1:]:
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ff, gg = f, g
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f = max(x, ff * x, gg * x)
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g = min(x, ff * x, gg * x)
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ans = max(ans, f)
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return ans
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```
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### **Java**
@@ -79,52 +78,14 @@ class Solution:
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```java
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class Solution {
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public int maxProduct(int[] nums) {
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int maxf = nums[0], minf = nums[0], res = nums[0];
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int f = nums[0], g = nums[0], ans = nums[0];
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for (int i = 1; i < nums.length; ++i) {
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int m = maxf, n = minf;
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maxf = Math.max(nums[i], Math.max(m * nums[i], n * nums[i]));
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minf = Math.min(nums[i], Math.min(m * nums[i], n * nums[i]));
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res = Math.max(res, maxf);
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}
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return res;
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}
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}
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```
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### **TypeScript**
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```ts
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function maxProduct(nums: number[]): number {
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let n = nums.length;
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let preMax = nums[0],
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preMin = nums[0],
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ans = nums[0];
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for (let i = 1; i < n; ++i) {
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let cur = nums[i];
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let x = preMax,
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y = preMin;
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preMax = Math.max(x * cur, y * cur, cur);
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preMin = Math.min(x * cur, y * cur, cur);
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ans = Math.max(preMax, ans);
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}
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return ans;
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}
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```
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### **C#**
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```cs
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public class Solution {
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public int MaxProduct(int[] nums) {
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int maxf = nums[0], minf = nums[0], res = nums[0];
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for (int i = 1; i < nums.Length; ++i)
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{
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int m = maxf, n = minf;
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maxf = Math.Max(nums[i], Math.Max(nums[i] * m, nums[i] * n));
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minf = Math.Min(nums[i], Math.Min(nums[i] * m, nums[i] * n));
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res = Math.Max(res, maxf);
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int ff = f, gg = g;
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f = Math.max(nums[i], Math.max(ff * nums[i], gg * nums[i]));
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g = Math.min(nums[i], Math.min(ff * nums[i], gg * nums[i]));
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ans = Math.max(ans, f);
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}
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return res;
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return ans;
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}
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}
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```
@@ -135,14 +96,14 @@ public class Solution {
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class Solution {
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public:
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int maxProduct(vector<int>& nums) {
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int maxf = nums[0], minf = nums[0], res = nums[0];
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int f = nums[0], g = nums[0], ans = nums[0];
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for (int i = 1; i < nums.size(); ++i) {
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int m = maxf, n = minf;
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maxf = max(nums[i], max(nums[i] * m, nums[i] * n));
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minf = min(nums[i], min(nums[i] * m, nums[i] * n));
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res = max(res, maxf);
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int ff = f, gg = g;
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f = max({nums[i], ff * nums[i], gg * nums[i]});
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g = min({nums[i], ff * nums[i], gg * nums[i]});
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ans = max(ans, f);
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}
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return res;
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return ans;
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}
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};
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```
@@ -151,14 +112,14 @@ public:
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```go
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func maxProduct(nums []int) int {
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maxf, minf, res := nums[0], nums[0], nums[0]
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for i := 1; i < len(nums); i++ {
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m, n := maxf, minf
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maxf = max(nums[i], max(nums[i]*m, nums[i]*n))
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minf = min(nums[i], min(nums[i]*m, nums[i]*n))
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res = max(res, maxf)
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f, g, ans := nums[0], nums[0], nums[0]
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for _, x := range nums[1:] {
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ff, gg := f, g
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f = max(x, max(ff*x, gg*x))
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g = min(x, min(ff*x, gg*x))
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ans = max(ans, f)
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}
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return res
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return ans
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}
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func max(a, b int) int {
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}
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```
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### **TypeScript**
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```ts
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function maxProduct(nums: number[]): number {
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let [f, g, ans] = [nums[0], nums[0], nums[0]];
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for (let i = 1; i < nums.length; ++i) {
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const [ff, gg] = [f, g];
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f = Math.max(nums[i], ff * nums[i], gg * nums[i]);
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g = Math.min(nums[i], ff * nums[i], gg * nums[i]);
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ans = Math.max(ans, f);
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}
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return ans;
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}
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```
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### **C#**
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```cs
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public class Solution {
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public int MaxProduct(int[] nums) {
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int f = nums[0], g = nums[0], ans = nums[0];
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for (int i = 1; i < nums.Length; ++i) {
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int ff = f, gg = g;
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f = Math.Max(nums[i], Math.Max(ff * nums[i], gg * nums[i]));
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g = Math.Min(nums[i], Math.Min(ff * nums[i], gg * nums[i]));
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ans = Math.Max(ans, f);
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}
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return ans;
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}
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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn max_product(nums: Vec<i32>) -> i32 {
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let mut min = nums[0];
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let mut max = nums[0];
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let mut res = nums[0];
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for &num in nums.iter().skip(1) {
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let (pre_min, pre_max) = (min, max);
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min = num.min(num * pre_min).min(num * pre_max);
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max = num.max(num * pre_min).max(num * pre_max);
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res = res.max(max);
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let mut f = nums[0];
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let mut g = nums[0];
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let mut ans = nums[0];
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for &x in nums.iter().skip(1) {
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let (ff, gg) = (f, g);
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f = x.max(x * ff).max(x * gg);
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g = x.min(x * ff).min(x * gg);
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ans = ans.max(f);
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}
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res
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ans
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}
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}
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```
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### **JavaScript**
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```js
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/**
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* @param {number[]} nums
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* @return {number}
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*/
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var maxProduct = function (nums) {
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let [f, g, ans] = [nums[0], nums[0], nums[0]];
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for (let i = 1; i < nums.length; ++i) {
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const [ff, gg] = [f, g];
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f = Math.max(nums[i], ff * nums[i], gg * nums[i]);
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g = Math.min(nums[i], ff * nums[i], gg * nums[i]);
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ans = Math.max(ans, f);
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}
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return ans;
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};
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```
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### **...**
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```

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