package jz.bm;import javax.crypto.MacSpi;
import java.util.ArrayList;
import java.util.Arrays;public class bm7 {/*** BM62 斐波那契数列*/public int Fibonacci(int n) {if (n <= 2) {return 1;}int[] dp = new int[n + 1];dp[1] = 1;dp[2] = 1;for (int i = 3; i <= n; i++) {dp[i] = dp[i - 1] + dp[i - 2];}return dp[n];}/*** BM63 跳台阶*/public int jumpFloor (int number) {int[] dp = new int[number + 1];dp[0] = 1;dp[1] = 1;for (int i = 2; i <= number; i++) {dp[i] = dp[i - 2] + dp[i - 1];}return dp[number];}/*** BM64 最小花费爬楼梯*/public int minCostClimbingStairs (int[] cost) {int[] dp = new int[cost.length + 1];
//        dp[1] = 0; //只有一层已经在顶楼了
//        dp[2] = 0; //两层的话选择从2楼开始for (int i = 2; i <= cost.length; i++) {dp[i] = Math.min(dp[i - 2] + cost[i - 2], dp[i - 1] + cost[i - 1]);}return dp[cost.length];}/*** BM65 最长公共子序列(二)*/public String LCS (String s1, String s2) {if (s1 == null || s1.equals("") || s2 == null || s2.equals("")) {return "";}int m = s1.length(), n = s2.length();String[][] dp = new String[m + 1][n + 1]; //dp[i][j]表示从左到右，当处理到s1的第i个元素和s2的第j个元素时的公共子序列for (int i = 0; i <= m; i++) {dp[i][0] = "";}for (int i = 0; i <= n; i++) {dp[0][i] = "";}for (int i = 1; i <= m; i++) {for (int j = 1; j <= n; j++) {if (s1.charAt(i - 1) == s2.charAt(j - 1)) {dp[i][j] = dp[i - 1][j - 1] + s1.charAt(i - 1);} else {//取长度较长的情况dp[i][j] = dp[i - 1][j].length() > dp[i][j - 1].length() ? dp[i - 1][j] : dp[i][j - 1];}}}return dp[m][n].equals("") ? "-1" : dp[m][n];}/*** BM66 最长公共子串*/public String LCS66 (String str1, String str2) {if (str1 == null || str1.equals("") || str2 == null || str2.equals("")) {return "";}int max = 0; //每段子串都是断开的，记录最长的那一个int maxIndex = 0;int m = str1.length(), n = str2.length();int[][] dp = new int[m + 1][n + 1]; //dp[i][j]表示从左到右，当处理到s1的第i个元素和s2的第j个元素时的公共字串长度for (int i = 1; i <= m; i++) {for (int j = 1; j <= n; j++) {if (str1.charAt(i - 1) == str2.charAt(j - 1)) {dp[i][j] = dp[i - 1][j - 1] + 1;if (dp[i][j] > max) {max = dp[i][j];maxIndex = i;}} else {dp[i][j] = 0; //无法构成字串}}}return str1.substring(maxIndex - max, maxIndex);}/*** BM67 不同路径的数目(一)*/public int uniquePaths (int m, int n) {int[][] dp = new int[m + 1][n + 1];for (int i = 1; i <= m; i++) {for (int j = 1; j <= n; j++) {//只有一行或一列时，只有一张方法if (i == 1) {dp[i][j] = 1;continue;}if (j == 1) {dp[i][j] = 1;continue;}//只能来自于上或者右dp[i][j] = dp[i - 1][j] + dp[i][j - 1];}}return dp[m][n];}/*** BM68 矩阵的最小路径和*/public int minPathSum (int[][] matrix) {int row = matrix.length;int col = matrix[0].length;int[][] dp = new int[row][col];//初值dp[0][0] = matrix[0][0];for (int i = 1; i < row; i++) {dp[i][0] = dp[i - 1][0] + matrix[i][0];}for (int i = 1; i < col; i++) {dp[0][i] = dp[0][i - 1] + matrix[0][i];}for (int i = 1; i < row; i++) {for (int j = 1; j < col; j++) {//选择值最小的路径dp[i][j] = Math.min(dp[i - 1][j] + matrix[i][j], dp[i][j] + matrix[i][j]);}}return dp[row - 1][col - 1];}/*** BM69 把数字翻译成字符串*/public int solve (String nums) {if (nums == null | "".equals(nums) || nums.startsWith("0")) {return 0;}int n = nums.length();
//        for (int i = 1; i < n; i++) {
//            if (nums.charAt(i) == '0' && !(nums.charAt(i - 1) == '1' || nums.charAt(i - 1) == '2')) {
//                return 0;
//            }
//        }int[] dp = new int[n + 1];
//        Arrays.fill(dp, 1);dp[0] = 1;  //默认一个才行dp[1] = nums.charAt(0) == '0' ? 0 : 1;for (int i = 2; i <= n; i++) {//无法独立编码也无法组合编码 00 或者 30 ，40 ...if(nums.charAt(i-1) == '0' && (nums.charAt(i-2) == '0' || nums.charAt(i-2) > '2')){return 0;//只能组合编码}else if(nums.charAt(i-1)=='0'){ //10,20dp[i] = dp[i-2];//只能独立编码}else if(nums.charAt(i-2)=='0' || nums.charAt(i-2)>'2' || nums.charAt(i-2)=='2'&& nums.charAt(i-1)>'6' ){ //0x,超过27dp[i] = dp[i-1];//两种编码方式都可以}else{dp[i] = dp[i-1]+dp[i-2];}
//            int num = (nums.charAt(i - 2) - '0') * 10 + nums.charAt(i - 1) - '0';
//            if ((num >= 11 && num <= 19) || (num >= 21 && num <= 26)) { //可以组合又可以单独译码
//                dp[i] = dp[i - 1] + dp[i - 2];
//            } else { //只能单独破译或者是10，20
//                dp[i] = dp[i - 1];
//            }}return dp[n];}/*** BM70 兑换零钱(一)*/public int minMoney (int[] arr, int aim) {int[] dp = new int[aim + 1];//初值，都设为最大值，假设货币值1元，则最大数为aim，因此可以设置为aim + 1Arrays.fill(dp, aim + 1);dp[0] = 0;for (int i = 1; i <= aim; i++) {//枚举所有的面值，找到最小值for (int j = 0; j < arr.length; j++) {if (i - arr[j] >= 0) { //当前面值可用dp[i] = Math.min(dp[i], dp[i - arr[j]] + 1);}}}return dp[aim] > aim ? -1 : dp[aim];}/*** BM71 最长上升子序列(一)*/public int LIS (int[] arr) {int m = arr.length;if (m == 0) {return 0;}int[] dp = new int[m]; //以i为结尾的最长子序列//初值都设置1，每个数字都是一个序列Arrays.fill(dp, 1);int res = 1;for (int i = 1; i < m; i++) {//遍历之前的所有数字，找到子序列for (int j = 0; j < i; j++) {//小于结尾数字，序列长度加1if (arr[i] > arr[j]) {dp[i] = Math.max(dp[i], dp[j] + 1);}}res = Math.max(dp[i], res);}return res;}/*** BM72 连续子数组的最大和*/public int FindGreatestSumOfSubArray (int[] array) {int n = array.length;int[] dp = new int[n + 1];dp[0] = 0;int max = array[0];for (int i = 1; i <= n; i++) {dp[i] = Math.max(dp[i - 1] + array[i - 1], array[i - 1]);max = Math.max(max, dp[i]);}return max;}/*** BM73 最长回文子串*/public int getLongestPalindrome (String A) {int maxlen = 1;for(int i = 0; i < A.length() - 1; i++)maxlen = Math.max(maxlen, Math.max(getPalindrome(A, i, i), getPalindrome(A, i, i + 1)));return maxlen;}public int getPalindrome(String A, int left, int right) {while (left >= 0 && right < A.length() && A.charAt(left) == A.charAt(right)) {left--;right++;}//注意：长度要减1return right - left - 1;}/*** BM74 数字字符串转化成IP地址*///递归回溯public ArrayList<String> restoreIpAddresses (String s) {ArrayList<String> res = new ArrayList<>();dfs(s, 0, res, new ArrayList<String>());return res;}private void dfs(String s, int begin, ArrayList<String> res, ArrayList<String> cur) {//找到4段IP，并且字符串遍历完毕System.out.println(cur.size());if (cur.size() == 4 && begin == s.length()) {res.add(cur.get(0) + "." + cur.get(1) + "." + cur.get(2) + "." + cur.get(3));return;}//最多需要3 * （4 - cur.size())数字,最少需要（4 - cur.size()数字if ((s.length() - begin > 3 * (4 - cur.size())) || (s.length() - begin < (4 - cur.size()))) {return;}//寻找1到3位数字，作为当前的ip段for (int i = begin; i < begin + 3 && i < s.length(); i++) {String s1 = s.substring(begin, i + 1);int num = Integer.parseInt(s1);//小于255且不能有前导0if (num >= 0 && num <= 255 && (s1.length() == 1 || s1.charAt(0) != '0')) {cur.add(s1);dfs(s,i + 1, res, cur);cur.remove(cur.size() - 1);}}}/*** BM75 编辑距离(一)*/public int editDistance (String str1, String str2) {int m = str1.length(), n = str2.length();int[][] dp = new int[m + 1][n + 1];//以i，j为结尾的字符串str1和str2，最少操作数//m为0时，str1变成str2最少操作数是str1插入字符for (int i = 1; i <= n; i++) {dp[0][i] = dp[0][i - 1] + 1;}//n为0时，str1变成str2最少操作数是str1删除字符for (int i = 1; i <= m; i++) {dp[i][0] = dp[i - 1][0] + 1;}for (int i = 1; i <= m; i++) {for (int j = 1; j <= n; j++) {//不需要额外操作if (str1.charAt(i - 1) == str2.charAt(j - 1)) {dp[i][j] = dp[i - 1][j - 1];} else {//插入、删除、修改都可以dp[i][j] = Math.min(dp[i - 1][j], Math.min(dp[i][j - 1], dp[i - 1][j - 1])) + 1;}}}return dp[m][n];}/*** BM78 打家劫舍(一)*/public int rob (int[] nums) {int n = nums.length;int[] dp = new int[n + 1];dp[1] = nums[0];for (int i = 2; i <= n; i++) {//偷这家，不偷这家dp[i] = Math.max(nums[i - 1] + dp[i - 2], dp[i - 1]);}return dp[n];}/*** BM79 打家劫舍(二)*/public int rob2 (int[] nums) {int n = nums.length;int[] dp = new int[n + 1];//必须要分情况了//偷第一家,最后一家不可以偷dp[1] = nums[0];for (int i = 2; i < n; i++) {//偷这家，不偷这家dp[i] = Math.max(nums[i - 1] + dp[i - 2], dp[i - 1]);}int res = dp[n - 1];//不偷第一家Arrays.fill(dp, 0);dp[1] = 0;for (int i = 2; i <= n; i++) {//偷这家，不偷这家dp[i] = Math.max(nums[i - 1] + dp[i - 2], dp[i - 1]);}return Math.max(res, dp[n]);}/*** BM80 买卖股票的最好时机(一)*/public int maxProfit (int[] prices) {int n = prices.length;if (n == 0) return 0;int res = 0;int min = prices[0];for (int i = 1; i < n; i++) {res = Math.max(prices[i] - min, res);min = Math.min(min, prices[i]);}return res;}/*** BM81 买卖股票的最好时机(二)*/public int maxProfit1 (int[] prices) {int n = prices.length;if (n <= 1) return 0;int res = 0;for (int i = 1; i < n; i++) {if (prices[i] - prices[i - 1] > 0) {res += prices[i] - prices[i - 1];}}return res;}/****/
}