第一题：1143. Longest Common Subsequence
/*二维dp数组
*/
class Solution {public int longestCommonSubsequence(String text1, String text2) {// char[] char1 = text1.toCharArray();// char[] char2 = text2.toCharArray();// 可以在一開始的時候就先把text1, text2 轉成char[]，之後就不需要有這麼多爲了處理字串的調整// 就可以和卡哥的code更一致int[][] dp = new int[text1.length() + 1][text2.length() + 1]; // 先对dp数组做初始化操作for (int i = 1 ; i <= text1.length() ; i++) {char char1 = text1.charAt(i - 1);for (int j = 1; j <= text2.length(); j++) {char char2 = text2.charAt(j - 1);if (char1 == char2) { // 开始列出状态转移方程dp[i][j] = dp[i - 1][j - 1] + 1;} else {dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);}}}return dp[text1.length()][text2.length()];}
}/**一维dp数组
*/
class Solution {public int longestCommonSubsequence(String text1, String text2) {int n1 = text1.length();int n2 = text2.length();// 多从二维dp数组过程分析 // 关键在于 如果记录 dp[i - 1][j - 1]// 因为 dp[i - 1][j - 1] <!=> dp[j - 1] <=> dp[i][j - 1]int [] dp = new int[n2 + 1];for(int i = 1; i <= n1; i++){// 这里pre相当于 dp[i - 1][j - 1]int pre = dp[0];for(int j = 1; j <= n2; j++){//用于给pre赋值int cur = dp[j];if(text1.charAt(i - 1) == text2.charAt(j - 1)){//这里pre相当于dp[i - 1][j - 1] 千万不能用dp[j - 1] !!dp[j] = pre + 1;} else{// dp[j] 相当于 dp[i - 1][j]// dp[j - 1] 相当于 dp[i][j - 1]dp[j] = Math.max(dp[j], dp[j - 1]);}//更新dp[i - 1][j - 1], 为下次使用做准备pre = cur;}}return dp[n2];}
}第二题：1035. Uncrossed Lines
class Solution {public int maxUncrossedLines(int[] nums1, int[] nums2) {int len1 = nums1.length;int len2 = nums2.length;int[][] dp = new int[len1 + 1][len2 + 1];for (int i = 1; i <= len1; i++) {for (int j = 1; j <= len2; j++) {if (nums1[i - 1] == nums2[j - 1]) {dp[i][j] = dp[i - 1][j - 1] + 1;} else {dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);}}}return dp[len1][len2];}
}第三题：53. Maximum Subarray
/*** 1.dp[i]代表当前下标对应的最大值* 2.递推公式 dp[i] = max (dp[i-1]+nums[i],nums[i]) res = max(res,dp[i])* 3.初始化 都为 0* 4.遍历方向，从前往后* 5.举例推导结果。。。** @param nums* @return*/public static int maxSubArray(int[] nums) {if (nums.length == 0) {return 0;}int res = nums[0];int[] dp = new int[nums.length];dp[0] = nums[0];for (int i = 1; i < nums.length; i++) {dp[i] = Math.max(dp[i - 1] + nums[i], nums[i]);res = res > dp[i] ? res : dp[i];}return res;}
//因为dp[i]的递推公式只与前一个值有关，所以可以用一个变量代替dp数组，空间复杂度为O(1)
class Solution {public int maxSubArray(int[] nums) {int res = nums[0];int pre = nums[0];for(int i = 1; i < nums.length; i++) {pre = Math.max(pre + nums[i], nums[i]);res = Math.max(res, pre);}return res;}
}第四题：392. Is Subsequence
class Solution {public boolean isSubsequence(String s, String t) {int length1 = s.length(); int length2 = t.length();int[][] dp = new int[length1+1][length2+1];for(int i = 1; i <= length1; i++){for(int j = 1; j <= length2; j++){if(s.charAt(i-1) == t.charAt(j-1)){dp[i][j] = dp[i-1][j-1] + 1;}else{dp[i][j] = dp[i][j-1];}}}if(dp[length1][length2] == length1){return true;}else{return false;}}
}
修改遍历顺序后，可以利用滚动数组，对dp数组进行压缩
class Solution {public boolean isSubsequence(String s, String t) {// 修改遍历顺序，外圈遍历t，内圈遍历s。使得dp的推算只依赖正上方和左上方，方便压缩。int[][] dp = new int[t.length() + 1][s.length() + 1];for (int i = 1; i < dp.length; i++) { // 遍历t字符串for (int j = 1; j < dp[i].length; j++) { // 遍历s字符串if (t.charAt(i - 1) == s.charAt(j - 1)) {dp[i][j] = dp[i - 1][j - 1] + 1;} else {dp[i][j] = dp[i - 1][j];}}System.out.println(Arrays.toString(dp[i]));}return dp[t.length()][s.length()] == s.length();}
}
状态压缩
class Solution {public boolean isSubsequence(String s, String t) {int[] dp = new int[s.length() + 1];for (int i = 0; i < t.length(); i ++) {// 需要使用上一轮的dp[j - 1]，所以使用倒序遍历for (int j = dp.length - 1; j > 0; j --) {// i遍历的是t字符串，j遍历的是dp数组，dp数组的长度比s的大1，因此需要减1。if (t.charAt(i) == s.charAt(j - 1)) {dp[j] = dp[j - 1] + 1;}}}return dp[s.length()] == s.length();}
}
将dp定义为boolean类型，dp[i]直接表示s.substring(0, i)是否为t的子序列
class Solution { public boolean isSubsequence(String s, String t) {boolean[] dp = new boolean[s.length() + 1];// 表示 “” 是t的子序列dp[0] = true;for (int i = 0; i < t.length(); i ++) {for (int j = dp.length - 1; j > 0; j --) {if (t.charAt(i) == s.charAt(j - 1)) {dp[j] = dp[j - 1];}}}return dp[dp.length - 1];}
}
