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[pull] master from youngyangyang04:master #297

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Merged
pull merged 6 commits into AlgorithmAndLeetCode:master from youngyangyang04:master
Jul 2, 2023
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10 changes: 5 additions & 5 deletions problems/0131.分割回文串.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -118,7 +118,7 @@ for (int i = startIndex; i < s.size(); i++) {
continue;
}
backtracking(s, i + 1); // 寻找i+1为起始位置的子串
path.pop_back(); // 回溯过程,弹出本次已经填在的子串
path.pop_back(); // 回溯过程,弹出本次已经添加的子串
}
```

Expand Down Expand Up @@ -189,7 +189,7 @@ private:
continue;
}
backtracking(s, i + 1); // 寻找i+1为起始位置的子串
path.pop_back(); // 回溯过程,弹出本次已经填在的子串
path.pop_back(); // 回溯过程,弹出本次已经添加的子串
}
}
bool isPalindrome(const string& s, int start, int end) {
Expand Down Expand Up @@ -245,7 +245,7 @@ private:
continue;
}
backtracking(s, i + 1); // 寻找i+1为起始位置的子串
path.pop_back(); // 回溯过程,弹出本次已经填在的子串
path.pop_back(); // 回溯过程,弹出本次已经添加的子串
}
}
void computePalindrome(const string& s) {
Expand Down Expand Up @@ -437,7 +437,7 @@ class Solution:
substring = s[startIndex:i + 1]
path.append(substring)
self.backtracking(s, i + 1, path, result, isPalindrome) # 寻找i+1为起始位置的子串
path.pop() # 回溯过程,弹出本次已经填在的子串
path.pop() # 回溯过程,弹出本次已经添加的子串

def computePalindrome(self, s, isPalindrome):
for i in range(len(s) - 1, -1, -1): # 需要倒序计算,保证在i行时,i+1行已经计算好了
Expand Down Expand Up @@ -497,7 +497,7 @@ func dfs(s string, start int) {
if isPalindrome(str) { // 是回文子串
path = append(path, str)
dfs(s, i+1) // 寻找i+1为起始位置的子串
path = path[:len(path)-1] // 回溯过程,弹出本次已经填在的子串
path = path[:len(path)-1] // 回溯过程,弹出本次已经添加的子串
}
}
}
Expand Down
37 changes: 36 additions & 1 deletion problems/0188.买卖股票的最佳时机IV.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -227,7 +227,7 @@ class Solution {
}
}

//版本三:一维 dp数组
//版本三:一维 dp数组 (下面有和卡哥邏輯一致的一維數組JAVA解法)
class Solution {
public int maxProfit(int k, int[] prices) {
if(prices.length == 0){
Expand Down Expand Up @@ -259,6 +259,41 @@ class Solution {
}
}
```
```JAVA
class Solution {
public int maxProfit(int k, int[] prices) {

//edge cases
if(prices.length == 0 || k == 0)
return 0;


int dp[] = new int [k * 2 + 1];

//和卡哥邏輯一致,奇數天購入股票,故初始化只初始化奇數天。
for(int i = 1; i < 2 * k + 1; i += 2){
dp[i] = -prices[0];
}

for(int i = 1; i < prices.length; i++){ //i 從 1 開始,因爲第 i = 0 天已經透過初始化完成了。
for(int j = 1; j < 2 * k + 1; j++){ //j 從 1 開始,因爲第 j = 0 天已經透過初始化完成了。
//奇數天購買
if(j % 2 == 1)
dp[j] = Math.max(dp[j], dp[j - 1] - prices[i]);
//偶數天賣出
else
dp[j] = Math.max(dp[j], dp[j - 1] + prices[i]);
}
//打印DP數組
//for(int x : dp)
// System.out.print(x +", ");
//System.out.println();
}
//return 第2 * k次賣出的獲利。
return dp[2 * k];
}
}
```

Python:

Expand Down
1 change: 1 addition & 0 deletions problems/0674.最长连续递增序列.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -177,6 +177,7 @@ Java:
dp[i] = 1;
}
int res = 1;
//可以注意到,這邊的 i 是從 0 開始,所以會出現和卡哥的C++ code有差異的地方,在一些地方會看到有 i + 1 的偏移。
for (int i = 0; i < nums.length - 1; i++) {
if (nums[i + 1] > nums[i]) {
dp[i + 1] = dp[i] + 1;
Expand Down

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