学习笔记 |#|Title|Solutions| |---|---|------| |1143|[longest-common-subsequence](https://leetcode-cn.com/problems/longest-common-subsequence) | 递归([Go](1143/longest_common_subsequence.go),[Py](1143/longest_common_subsequence.py)),动态规划([Go](1143/longest_common_subsequence2.go),[Py](1143/longest_common_subsequence2.py)),动态规划2([Go](1143/longest_common_subsequence3.go),[Py](1143/longest_common_subsequence3.py))| |213|[house-robber-ii](https://leetcode-cn.com/problems/house-robber-ii) | 动态规划([Go](213/house_robber_ii.go),[Py](213/house_robber_ii.py))| |62|[unique-paths](https://leetcode-cn.com/problems/unique-paths) | 动态规划([Go](62/unique_paths.go),[Py](62/unique_paths.py))| |63|[unique-paths-ii](https://leetcode-cn.com/problems/unique-paths-ii) | 动态规划([Go](63/unique_path_ii.go),[Py](63/unique_path_ii.py))| |120|[triangle](https://leetcode-cn.com/problems/triangle) | 动态规划([Go](120/triangle.go),[Py](120/triangle.py))| |53|[maximum-subarray](https://leetcode-cn.com/problems/maximum-subarray) | 动态规划([Go](../Week_06/53/maximum_subarray.go),[Py](../Week_06/53/maximum_subarray.py)),分治([Go](../Week_06/53/maximum_subarray2.go),[Py](../Week_06/53/maximum_subarray2.py))| |152|[maximum-product-subarray](https://leetcode-cn.com/problems/maximum-product-subarray) | 动态规划([Go](../Week_06/152/maximum_product_subarray.go),[Py](../Week_06/152/maximum_product_subarray.py))| |322|[coin-change](https://leetcode-cn.com/problems/coin-change) | 动态规划([Go](322/coin_change.go),[Py](322/coin_change.py))| |198|[house-robber](https://leetcode-cn.com/problems/house-robber) | 动态规划([Go](198/house_robber.go),[Py](198/house_robber.py))| |121|[best-time-to-buy-and-sell-stock](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock) | 动态规划([Go](121/best_time_to_buy_and_sell_stock.go),[Py](121/best_time_to_buy_and_sell_stock.py))| |91|[decode-ways](https://leetcode-cn.com/problems/decode-ways) | 动态规划([Go](91/decode_ways.go),[Py](91/decode_ways.py))| |123|[best-time-to-buy-and-sell-stock-iii](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-iii) | 动态规划([Go](123/best_time_to_buy_and_sell_stock_iii.go),[Py](123/best_time_to_buy_and_sell_stock_iii.py)),动态规划2([Go](123/best_time_to_buy_and_sell_stock_iii_2.go),[Py](123/best_time_to_buy_and_sell_stock_iii_2.py))| |309|[best-time-to-buy-and-sell-stock-with-cooldown](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-with-cooldown) | 动态规划([Go](309/best_time_to_buy_and_sell_stock_with_cooldown.go),[Py](309/best_time_to_buy_and_sell_stock_with_cooldown.py))| |188|[best-time-to-buy-and-sell-stock-iv](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-iv) | 动态规划([Go](188/best_time_to_buy_and_sell_stock_iv_2.go),[Py](188/best_time_to_buy_and_sell_stock_iv_2.py))| |714|[best-time-to-buy-and-sell-stock-with-transaction-fee](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee) | 动态规划([Go](714/best_time_to_buy_and_sell_stock_with_transaction_fee.go),[Py](714/best_time_to_buy_and_sell_stock_with_transaction_fee.py))| |279|[perfect-squares](https://leetcode-cn.com/problems/perfect-squares) | 动态规划([Go](279/perfect_squares.go),[Py](279/perfect_squares.py))| |518|[coin-change-2](https://leetcode-cn.com/problems/coin-change-2) | 动态规划([Go](518/coin_change_2.go),[Py](518/coin_change_2.py))| |221|[maximal-square](https://leetcode-cn.com/problems/maximal-square) | 动态规划([Go](221/maximal_square.go),[Py](221/maximal_square.py))| |621|[task-scheduler](https://leetcode-cn.com/problems/task-scheduler) | 数组([Go](621/task_scheduler.go),[Py](621/task_scheduler.py))| |647|[palindromic-substrings](https://leetcode-cn.com/problems/palindromic-substrings) | 动态规划([Go](647/palindromic_substrings.go),[Py](647/palindromic_substrings.py)), 中心扩展([Go](647/palindromic_substrings2.go),[Py](647/palindromic_substrings2.py))| |403|[frog-jump](https://leetcode-cn.com/problems/frog-jump) | 深度优先搜索([Go](403/frog_jump.go),[Py](403/frog_jump.py)), 动态规划([Go](403/frog_jump2.go),[Py](403/frog_jump2.py))| ## 题解 ### 1143. longest-common-subsequence 1. 递归 2. 动态规划: - dp[i][j] = dp[i-1][j-1] + 1 或 max(dp[i][j-1], dp[i-1][j]) 3. 动态规划空间优化: - 用一个变量pre保存dp[i-1][j-1]可以实现空间负载度降维 - dp[j] = pre+1 或者 max(dp[j], dp[j-1]) ### 62. unique-paths 1. 动态规划: - dp[i][j] = 1 if (i==0 or j ==0) else dp[i-1][j] + dp[i][j-1] ### 120. triangle 1. 动态规划: 倒序递推: ### 53. maximum-subarray 1. 动态规划: dp[i] = nums[i] + max(0, dp[i-1]) 2. 分治: ### 152. maximum-product-subarray 1. 动态规划: 用两个变量分别记录当前最小乘积和最大乘积 ### 322. coin_change 1. 动态规划: 建立一个极值数组: dp = [amount+1] *(amount+1); dp[i] = min(dp[i], dp[i-c]+1) ### 198. house-robber 1. 动态规划: ### 213. house-robber-ii 1. 动态规划: - 两次动态规划dp(nums[:-1])、dp(nums[1:]),取两者之间的较大值 ### 91. decode-ways 1. 动态规划: 首先需要将这道题类比爬楼梯问题，可能只能走两步，可能只能走一步，可能一步两步都可能走 - 特殊处理s[i] == '0' - 如果s[i-1] in ('1', '2')， 即走两步: dp[i+1] = dp[i-1] - 否则不合法 - 当s[i-1] == '1' 或者 (s[i-1] == '2' && '1' <= s[i] <= '6')， 即走一步或两步: dp[i+1] = dp[i] + dp[i-1] - 否则走一步 dp[i+1] = dp[i] ### 123. best-time-to-buy-and-sell-stock-iii 1. 动态规划: - 定义一个三维数组：第i天第j次买(0)或者卖(1) - 递推公式 - 第0天 dp[i][j][1] = -prices[i] - 第1~n天 - dp[i][j][1] = max(dp[i-1][j][1], dp[i-1][j-1][0]-prices[i]) - dp[i][j][0] = max(dp[i-1][j][0], dp[i-1][j][1]+prices[i]) 2. 动态规划二： 因为最多只有两次交易，所以只需要定义四个变量 dp11(第一次买入)、dp10(第一次卖出)、dp21(第二次买入)、dp20(第二次卖出) - 第0天： - dp11 = -prices[0] - dp10 = 0 - dp21 = -prices[0] （当做第0天买入两次卖出一次） - dp20 = 0 （当做第0天买入两次卖出两次） - 第1~n天: - dp11 = max(dp11, -prices[i]) - dp10 = max(dp10, dp11+prices[i]) - dp21 = max(dp21, dp10-prices[i]) - dp20 = max(dp20, dp21+prices[i]) ### 309. best-time-to-buy-and-sell-stock-with-cooldown 1. 动态规划: - 定义一个二维数：第i天是否买(1)或卖(0) - 递推 - 第0，1天 - dp[0][1] = -prices[0] - dp[1][0] = max(dp[0][0], dp[0][1]+prices[1]) - dp[1][1] = max(dp[0][1], dp[0][0]-prices[1]) - 第2~n天 - dp[i][1] = max(dp[i-1][1], dp[i-2][0]-prices[i]) - dp[i][0] = max(dp[i-1][0], dp[i-1][1]+prices[i]) ### 309. best-time-to-buy-and-sell-stock-iv 1. 动态规划: #123的通用情况 ### 714. best-time-to-buy-and-sell-stock-with-transaction-fee 1. 动态规划: 每一天只有持有和不持有两个状态，且只和前一天有关,故可以用两个变量hold和not_hold即可 - not_hold = max(not_hold, hold+prices[i]) - hold = max(hold, not_hold-prices[i]-fee) ### 279. perfect-squares 1. 动态规划: 类似于换硬币问题 ### 518. coin-change-2 1. 动态规划 ### 221. maximal-square 1. 动态规划: dp[i][j]代表以(i,j)为右下角的正方形边长 - i == 0 or j == 0: dp[i][j] = int(matrix[i][j]) - dp[i][j] = min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1]) + 1 ### 621. task-scheduler 1. 数组: - 用一个长度为26的数组counter记录每种task的次数 - 排序找到最大次数的task: counter[-1] - 则 res = (counter[-1]-1) * (n+1) + 1 + x (x为其他的task出现次数也等于counter[-1]) - 当任务种类比较多时可能出现res < len(tasks), 这是结果为len(tasks) ### 647. palindromic-substrings 1. 动态规划: dp[i][j]代表[i,j]区间的子串是否是回文子串：s[i] == s[j] && (j - i < 2 || dp[i + 1][j - 1]) 2. 中心扩展: 长度为n的字符串会生成2n-1个中心点，其中l=i/2, r == l+ i%2 ### 403. frog-jump 1. 深度优先 2. 动态规划 - dp[i][k] = dp[j][k] || dp[j][k-1] || dp[j][k+1]; k = stones[i]-stones[j],k必定<=j+1