diff --git a/notes/Leetcode 题解.md b/notes/Leetcode 题解.md index 1751dd5d..0d5defcb 100644 --- a/notes/Leetcode 题解.md +++ b/notes/Leetcode 题解.md @@ -2327,15 +2327,17 @@ Note that different sequences are counted as different combinations. Therefore the output is 7. ``` +完全背包。 + ```java public int combinationSum4(int[] nums, int target) { if (nums == null || nums.length == 0) return 0; int[] dp = new int[target + 1]; dp[0] = 1; for (int i = 1; i <= target; i++) { - for (int j = 0; j < nums.length; j++) { - if (nums[j] <= i) { - dp[i] += dp[i - nums[j]]; + for (int num : nums) { + if (num <= i) { + dp[i] += dp[i - num]; } } } @@ -2343,30 +2345,13 @@ public int combinationSum4(int[] nums, int target) { } ``` -**只能进行两次的股票交易** - -[Leetcode : 123. Best Time to Buy and Sell Stock III (Hard)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/description/) - -```java -public int maxProfit(int[] prices) { - int firstBuy = Integer.MIN_VALUE, firstSell = 0; - int secondBuy = Integer.MIN_VALUE, secondSell = 0; - for (int curPrice : prices) { - if (firstBuy < -curPrice) firstBuy = -curPrice; - if (firstSell < firstBuy + curPrice) firstSell = firstBuy + curPrice; - if (secondBuy < firstSell - curPrice) secondBuy = firstSell - curPrice; - if (secondSell < secondBuy + curPrice) secondSell = secondBuy + curPrice; - } - return secondSell; -} -``` - **只能进行 k 次的股票交易** [Leetcode : 188. Best Time to Buy and Sell Stock IV (Hard)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/description/) ```html -dp[i, j] = max(dp[i, j-1], prices[j] - prices[jj] + dp[i-1, jj]) { jj in range of [0, j-1] } = max(dp[i, j-1], prices[j] + max(dp[i-1, jj] - prices[jj])) +dp[i, j] = max(dp[i, j-1], prices[j] - prices[jj] + dp[i-1, jj]) { jj in range of [0, j-1] } + = max(dp[i, j-1], prices[j] + max(dp[i-1, jj] - prices[jj])) ``` ```java @@ -2392,6 +2377,24 @@ public int maxProfit(int k, int[] prices) { } ``` +**只能进行两次的股票交易** + +[Leetcode : 123. Best Time to Buy and Sell Stock III (Hard)](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/description/) + +```java +public int maxProfit(int[] prices) { + int firstBuy = Integer.MIN_VALUE, firstSell = 0; + int secondBuy = Integer.MIN_VALUE, secondSell = 0; + for (int curPrice : prices) { + if (firstBuy < -curPrice) firstBuy = -curPrice; + if (firstSell < firstBuy + curPrice) firstSell = firstBuy + curPrice; + if (secondBuy < firstSell - curPrice) secondBuy = firstSell - curPrice; + if (secondSell < secondBuy + curPrice) secondSell = secondBuy + curPrice; + } + return secondSell; +} +``` + ### 数组区间 **数组区间和** @@ -2824,8 +2827,7 @@ public int countPrimes(int n) { ```java int gcd(int a, int b) { - if (b == 0) return a; - return gcd(b, a % b); + return b == 0 ? a : gcd(b, a% b); } ``` @@ -2841,47 +2843,80 @@ int lcm(int a, int b){ 对于 a 和 b 的最大公约数 f(a, b)，有： -1\. 如果 a 和 b 均为偶数，f(a, b) = 2\*f(a/2, b/2); -2\. 如果 a 是偶数 b 是奇数，f(a, b) = f(a/2, b); -3\. 如果 b 是偶数 a 是奇数，f(a, b) = f(a, b/2); -4\. 如果 a 和 b 均为奇数，f(a, b) = f(a, a-b); +- 如果 a 和 b 均为偶数，f(a, b) = 2\*f(a/2, b/2); +- 如果 a 是偶数 b 是奇数，f(a, b) = f(a/2, b); +- 如果 b 是偶数 a 是奇数，f(a, b) = f(a, b/2); +- 如果 a 和 b 均为奇数，f(a, b) = f(a, a-b); 乘 2 和除 2 都可以转换为移位操作。 ### 进制转换 -Java 中 static String toString(int num, int radix) 可以将一个整数装换为 redix 进制表示的字符串。 - **7 进制** [Leetcode : 504. Base 7 (Easy)](https://leetcode.com/problems/base-7/description/) ```java public String convertToBase7(int num) { - if (num < 0) { - return '-' + convertToBase7(-num); - } - if (num < 7) { - return num + ""; - } + if (num < 0) return '-' + convertToBase7(-num); + if (num < 7) return num + ""; return convertToBase7(num / 7) + num % 7; } ``` +```java +public String convertToBase7(int num) { + if (num == 0) return "0"; + StringBuilder sb = new StringBuilder(); + boolean isNegative = num < 0; + if (isNegative) num = -num; + while (num > 0) { + sb.append(num % 7); + num /= 7; + } + String ret = sb.reverse().toString(); + return isNegative ? "-" + ret : ret; +} +``` + +Java 中 static String toString(int num, int radix) 可以将一个整数转换为 redix 进制表示的字符串。 + +```java +public String convertToBase7(int num) { + return Integer.toString(num, 7); +} +``` + **16 进制** [Leetcode : 405. Convert a Number to Hexadecimal (Easy)](https://leetcode.com/problems/convert-a-number-to-hexadecimal/description/) +负数要用它的补码形式。 + +```html +Input: +26 + +Output: +"1a" + +Input: +-1 + +Output: +"ffffffff" +``` + ```java public String toHex(int num) { char[] map = {'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'}; if(num == 0) return "0"; - String ret = ""; + StringBuilder sb = new StringBuilder(); while(num != 0){ - ret = map[(num & 0b1111)] + ret; - num >>>= 4; + sb.append(map[num & 0b1111]); + num >>>= 4; // 无符号右移，左边填 0 } - return ret; + return sb.reverse().toString(); } ``` @@ -2918,15 +2953,14 @@ Return "100". ```java public String addBinary(String a, String b) { int i = a.length() - 1, j = b.length() - 1, carry = 0; - String str = ""; - while(i >= 0 || j >= 0){ - if(i >= 0 && a.charAt(i--) == '1') carry++; - if(j >= 0 && b.charAt(j--) == '1') carry++; - str = (carry % 2) + str; + StringBuilder str = new StringBuilder(); + while (carry == 1 || i >= 0 || j >= 0) { + if (i >= 0 && a.charAt(i--) == '1') carry++; + if (j >= 0 && b.charAt(j--) == '1') carry++; + str.append(carry % 2); carry /= 2; } - if(carry == 1) str = "1" + str; - return str; + return str.reverse().toString(); } ``` @@ -2938,15 +2972,15 @@ public String addBinary(String a, String b) { ```java public String addStrings(String num1, String num2) { - StringBuilder sb = new StringBuilder(); - int carry = 0; - for(int i = num1.length() - 1, j = num2.length() - 1; i >= 0 || j >= 0 || carry == 1; i--, j--){ - int x = i < 0 ? 0 : num1.charAt(i) - '0'; - int y = j < 0 ? 0 : num2.charAt(j) - '0'; - sb.append((x + y + carry) % 10); + StringBuilder str = new StringBuilder(); + int carry = 0, i = num1.length() - 1, j = num2.length() - 1; + while (carry == 1 || i >= 0 || j >= 0) { + int x = i < 0 ? 0 : num1.charAt(i--) - '0'; + int y = j < 0 ? 0 : num2.charAt(j--) - '0'; + str.append((x + y + carry) % 10); carry = (x + y + carry) / 10; } - return sb.reverse().toString(); + return str.reverse().toString(); } ```