041、子序列类型问题（labuladong）

子序列类型问题

一、经典动态规划：编辑距离

基于labuladong的算法网站，经典动态规划：编辑距离；

总结：

• 一般来说涉及到两个字符串的问题，需要依赖上一次的各种操作，一般使用dp table；
• dp数组和dp函数：
  • 本质上是一样的；
  • 只不过dp数组是利用数组去体现结果值；
  • dp函数是在函数返回结果中体现；

[72]编辑距离//leetcode submit region begin(Prohibit modification and deletion)
class Solution {// 利用 dp table 进行自下而上的求解public int minDistance(String word1, String word2) {int m = word1.length();int n = word2.length();// dp[i][j] ： word1中从[0,i] 变成 word2[0,j] 所需要的最少步骤int[][] dp = new int[m + 1][n + 1];// base casefor (int i = 1; i <= m; i++) {dp[i][0] = i;}for (int i = 1; i <= n; i++) {dp[0][i] = i;}// 开始遍历for (int i = 1; i <= m; i++) {for (int j = 1; j <= n; j++) {// 判断if (word1.charAt(i - 1) == word2.charAt(j - 1)) {dp[i][j] = dp[i - 1][j - 1];} else {// 插入、删除、替换dp[i][j] = min(dp[i][j - 1] + 1,// 添加dp[i - 1][j] + 1,// 删除dp[i - 1][j - 1] + 1// 替换);}}}return dp[m][n];}int min(int a, int b, int c) {return Math.min(a, Math.min(b, c));}
}
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二、动态规划设计：最大子数组

基于labuladong的算法网站，动态规划设计：最大子数组；

力扣第53题，最大子数组和；

[53]最大子数组和//leetcode submit region begin(Prohibit modification and deletion)
class Solution {// dp tablepublic int maxSubArray(int[] nums) {int length = nums.length;// 需要明确dp数组的定义，dp[i]：以nums[i]为结尾时，最大连续子数组的最大和int[] dp = new int[length];// base casedp[0] = nums[0];for (int i = 1; i < length; i++) {dp[i] = Math.max(nums[i], dp[i - 1] + nums[i]);}// 遍历得到最大值int res = Integer.MIN_VALUE;for (int i = 0; i < length; i++) {if (dp[i] > res) {res = dp[i];}}return res;}
}
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三、经典动态规划：最长公共子序列

基于labuladong的算法网站，经典动态规划：最长公共子序列；

1、最长公共子序列

力扣第1143题，最长公共子序列；

[1143]最长公共子序列//leetcode submit region begin(Prohibit modification and deletion)
class Solution {// dp tablepublic int longestCommonSubsequence(String text1, String text2) {int m = text1.length();int n = text2.length();// dp[i][j] 代表字符串 text1[0,i] 和 text2[0,j] 的最长公共子序列长度int[][] dp = new int[m + 1][n + 1];// base casefor (int i = 0; i <= m; i++) {dp[m][0] = 0;}for (int i = 0; i <= n; i++) {dp[0][n] = 0;}// 开始遍历子串for (int i = 1; i <= m; i++) {for (int j = 1; j <= n; j++) {if (text1.charAt(i - 1) == text2.charAt(j - 1)) {dp[i][j] = dp[i - 1][j - 1] + 1;} else {dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);}}}// 返回return dp[m][n];}int max(int a, int b, int c) {return Math.max(a, Math.max(b, c));}
}
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2、两个字符串的删除操作

力扣第583题，两个字符串的删除操作;

[583]两个字符串的删除操作//leetcode submit region begin(Prohibit modification and deletion)
class Solution {/*** @param word1* @param word2* @return 返回使得 word1 和 word2 相同所需的最小步数*/public int minDistance(String word1, String word2) {int length = get(word1, word2);return word1.length() + word2.length() - length - length;}// 求 word1 和 word2 的最长公共子序列int get(String word1, String word2) {int m = word1.length();int n = word2.length();int[][] dp = new int[m + 1][n + 1];for (int i = 0; i <= m; i++) {dp[m][0] = 0;}for (int i = 0; i <= n; i++) {dp[0][n] = 0;}for (int i = 1; i <= m; i++) {for (int j = 1; j <= n; j++) {if (word1.charAt(i - 1) == word2.charAt(j - 1)) {dp[i][j] = dp[i - 1][j - 1] + 1;} else {dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);}}}return dp[m][n];}
}
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3、两个字符串的最小ASCII删除和

力扣第712题，两个字符串的最小ASCII删除和；

[712]两个字符串的最小ASCII删除和//leetcode submit region begin(Prohibit modification and deletion)
class Solution {public int minimumDeleteSum(String s1, String s2) {int m = s1.length();int n = s2.length();// dp[i][j] 使字符串 s1[0,i] 和 s2[0,j] 两个相等所需要删除的字符的 ASCII 值的最小和int[][] dp = new int[m + 1][n + 1];// base casefor (int i = 1; i <= m; i++) {dp[i][0] = dp[i - 1][0] + s1.charAt(i - 1);}for (int i = 1; i <= n; i++) {dp[0][i] = dp[0][i - 1] + s2.charAt(i - 1);}// 开始遍历for (int i = 1; i <= m; i++) {int code1 = s1.charAt(i - 1);for (int j = 1; j <= n; j++) {int code2 = s2.charAt(j - 1);// 判断此时两个字母是否相等if (code1 == code2) {dp[i][j] = dp[i - 1][j - 1];} else {dp[i][j] = Math.min(dp[i - 1][j] + code1, dp[i][j - 1] + code2);}}}return dp[m][n];}
}
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四、动态规划之子序列问题解题模板

基于labuladong的算法网站，动态规划之子序列问题解题模板；

1、最长回文子序列

力扣第516题，最长回文子序列；

[516]最长回文子序列//leetcode submit region begin(Prohibit modification and deletion)
class Solution {// 可以看成两个字符串，找最长公共子序列的长度public int longestPalindromeSubseq(String s) {StringBuffer sb = new StringBuffer();int length = s.length();for (int i = length - 1; i >= 0; i--) {sb.append(s.charAt(i));}return get(s, sb.toString());}int get(String word1, String word2) {int m = word1.length();int n = word2.length();int[][] dp = new int[m + 1][n + 1];for (int i = 1; i <= m; i++) {for (int j = 1; j <= n; j++) {if (word1.charAt(i - 1) == word2.charAt(j - 1)) {dp[i][j] = dp[i - 1][j - 1] + 1;} else {dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);}}}return dp[m][n];}
}
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2、让字符串成为回文串的最少插入次数

力扣第1312题，让字符串成为回文串的最少插入次数；

[1312]让字符串成为回文串的最少插入次数//leetcode submit region begin(Prohibit modification and deletion)
class Solution {public int minInsertions(String s) {StringBuffer sb = new StringBuffer();int length = s.length();for (int i = length - 1; i >= 0; i--) {sb.append(s.charAt(i));}int max = get(s, sb.toString());return length - max;}int get(String word1, String word2) {int m = word1.length();int n = word2.length();int[][] dp = new int[m + 1][n + 1];for (int i = 1; i <= m; i++) {for (int j = 1; j <= n; j++) {if (word1.charAt(i - 1) == word2.charAt(j - 1)) {dp[i][j] = dp[i - 1][j - 1] + 1;} else {dp[i][j] = Math.max(dp[i][j - 1], dp[i - 1][j]);}}}return dp[m][n];}
}
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