春招冲刺百题计划|栈

Java基础复习

Java数组的声明与初始化
Java ArrayList
Java HashMap
Java String 类
Java LinkedList
Java Deque继承LinkedList
Java Set

第一题：有效的括号

很简单的题，从大一做到现在，就是复习一下语法。
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class Solution {public boolean isValid(String s) {//先明确一下Java的stack用什么：Deque<Character> stack = new LinkedList<Character>();先进后出为栈//自己可以规定一下用法：addLast, removeLast和getLastint n = s.length();if(n%2!=0){return false;}Deque<Character> stack = new LinkedList<Character>();for(int i=0; i<n; i++){char cur = s.charAt(i);if(cur=='('||cur=='{'||cur=='['){stack.addLast(cur);continue;}if(stack.size()!=0){char top = stack.getLast();if(top=='(' && cur==')'){stack.removeLast();continue;}if(top=='{'&&cur=='}'){stack.removeLast();continue;}if(top=='['&&cur==']'){stack.removeLast();continue;}}return false;   }if(stack.size()==0){return true;}else{return false;}}
}

第二题：二叉树的中序遍历

中序遍历顺序：左根右。
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/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {//怎么改进成用栈呢？这里学习一位大佬的颜色标记法。作者：henrypublic List<Integer> inorderTraversal(TreeNode root) {List<Integer> res = new ArrayList<>();Deque<Object> stack = new LinkedList<>();if(root == null){return res;}stack.push("WHITE");stack.push(root);while(!stack.isEmpty()){TreeNode node = (TreeNode)stack.pop();String color = (String)stack.pop();if (node == null) continue;if(color.equals("WHITE")){stack.push("WHITE");stack.push(node.right);stack.push("GRAY");stack.push(node);stack.push("WHITE");stack.push(node.left);}else{res.add(node.val);}}return res;}//非常简单的回溯法// List<Integer> res = new ArrayList<>();// private void traceBack(TreeNode cur){//     if(cur.left != null){//        traceBack(cur.left);//     }//     res.add(cur.val);//     if(cur.right != null){//         traceBack(cur.right);//     }// }// public List<Integer> inorderTraversal(TreeNode root) {//     if(root != null){//         traceBack(root);//     }//     return res;// }
}

第三题：移掉k个数字，要求返回最小

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维持一个单调栈，单调栈最麻烦的是维持个数。在大佬的题解里面就直接舍弃这个。最后再处理。很有趣。
强推这个大佬的题解！一招吃遍力扣四道题，妈妈再也不用担心我被套路啦～

class Solution {public String removeKdigits(String num, int k) {//特殊情况：k>=num.length()int n = num.length();int remain = n-k;if(remain<=0){return "0";}//定义单调栈Deque<Character> stack = new LinkedList<Character>();//遍历整个num，在其中维护一个长度为无限制的单调栈for(int i=0; i<n; i++){char ch = num.charAt(i);while(!stack.isEmpty()&&k>0&&stack.peekLast()>ch){stack.pollLast();k--;}stack.offerLast(ch);}//只保留栈底的remain个元素for (int i = 0; i < k; ++i) {stack.pollLast();}//处理前导0StringBuilder ret = new StringBuilder();boolean leadingZero = true;while (!stack.isEmpty()) {char digit = stack.pollFirst();if (leadingZero && digit == '0') {continue;}leadingZero = false;ret.append(digit);}return ret.length() == 0 ? "0" : ret.toString();}// int mul[];// int n;// int res=Integer.MAX_VALUE;// int len;// List<Integer> path = new ArrayList<>();// private void traceBack(String num, int idx){//     if(idx>n){//         return;//     }//    // System.out.println("path= "+path);//     if(path.size()==len){//         int tmp=0;   //         for(int i=0; i<len; i++){//             tmp += mul[i] * path.get(i);//         }//         res = Math.min(tmp, res);//         // System.out.println(res + " "+ tmp);//         return;//     }//     for(int i=idx; i<n; i++){//         path.add(num.charAt(i)-'0');//         traceBack(num, i+1);//         path.remove(path.size()-1);//     }// }// public String removeKdigits(String num, int k) {//     //特殊情况：k>=num.length()//     n = num.length();//     if(k >= n){//         return "0";//     }//     len = n-k;//     mul = new int[n-k];//     for(int i=0; i<n-k; i++){//         mul[i] = (int) Math.pow((int)10, n-k-i-1);//         //System.out.println(mul[i]);//     }//     //肯定是把k全部用完最好。那就不删除，选择组合？变成组合问题。又想递归回溯了。//     //如果是回溯的话，我要画递归树，解空间（i之后的所有）深度（n-k）(不出所料，超时了。)//     traceBack(num, 0);//     return res+"";//     //其实我心里有一个想法，固定栈，里面保证是当前遍历的最小的四个。好像在哪里做过。我需要清醒的脑子过一遍，明天早上来。//     //起床了，还是想不到；看了题解：贪心+单调栈// }
}

第四题：去除重复字母，要求返回字典序最小

与上一题基本一致！
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class Solution {public String removeDuplicateLetters(String s) {//看过题解过后写的，与402的核心思路是相同的。//前期工作，得到一个map (字母：出现的次数)Map<Character, Integer> container = new HashMap<>();for(int i=0; i<s.length(); i++){if(!container.containsKey(s.charAt(i))){container.put(s.charAt(i), 1);}else{container.put(s.charAt(i), container.get(s.charAt(i))+1);}}//现在出现一个问题，就是前面保留了某个字母之后，后面怎么知道呢？题解里面由加了一个set:当已经保留了，之后的就不在讨论，且数目-1Set<Character> seen = new HashSet<>();int k = container.size();Deque<Character> stack = new LinkedList<>();for(int i=0; i<s.length(); i++){if(!seen.contains(s.charAt(i))){while(!stack.isEmpty()&&container.get(stack.getLast())>1&&stack.getLast()>=s.charAt(i)){container.put(stack.getLast(), container.get(stack.getLast())-1);seen.remove(stack.getLast());stack.removeLast();} stack.addLast(s.charAt(i));seen.add(s.charAt(i));System.out.print(container + "  ");System.out.println(stack);}else{container.put(s.charAt(i), container.get(s.charAt(i))-1);}}//只取满足条件的前k个。String res="";while(!stack.isEmpty()&&k>0){res += stack.getFirst()+"";stack.removeFirst();k--;}return res;}
}

第五题：拼接最大数，与前面两题相比，多了一个合并排序。

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class Solution {private int[] maxNumberInOne(int[] nums, int k){int res[] = new int[k];int remain = nums.length - k;Deque<Integer> stack = new LinkedList<>();for(int i=0; i<nums.length; i++){while(!stack.isEmpty()&&stack.getLast()<nums[i]&&remain>0){stack.removeLast();remain--;}stack.addLast(nums[i]);}for(int i=0; i<k&&!stack.isEmpty(); i++){res[i] = stack.getFirst();stack.removeFirst();}return res;}//不是简单的归并：查看官方题解，写了一个compare函数比如：public int compare(int[] subsequence1, int index1, int[] subsequence2, int index2) {int x = subsequence1.length, y = subsequence2.length;//按每一位比较while (index1 < x && index2 < y) {int difference = subsequence1[index1] - subsequence2[index2];if (difference != 0) {return difference;}index1++;index2++;}//比较剩余长度 谁的剩余长度大，谁就大return (x - index1) - (y - index2);}private int[] merge(int[] nums1, int[] nums2){int n1 = nums1.length;int n2 = nums2.length;int res[] = new int[n1+n2];Deque<Integer> stack = new LinkedList<>();Arrays.fill(res, 0);int i=0, j=0, k=0;for(; k<n1+n2; k++){if(compare(nums1, i, nums2, j)>0){stack.addLast(nums1[i]);i++;}else{stack.addLast(nums2[j]);j++;}res[k] = stack.getFirst();stack.removeFirst();}return res;}public int[] maxNumber(int[] nums1, int[] nums2, int k) {int res[] = new int[k];Arrays.fill(res, 0);//遍历k1和k2的多种可能(考虑挺多的):1.对于一个数组，最少选多少，最多选多少int n1 = nums1.length;int n2 = nums2.length;int maxk=Math.min(k, n1);int mink;if(n2-k>=0){mink = 0;}else{mink = k-n2; }for(int k1=mink; k1<=maxk; k1++){// System.out.println("min:" + k1 + ", max:" + (k-k1));int res1[] = maxNumberInOne(nums1, k1);int res2[] = maxNumberInOne(nums2, k-k1); // System.out.println("1: "+ Arrays.toString(res1) + "; 2: "+ Arrays.toString(res2));int tmp[] = merge(res1, res2);// System.out.println( Arrays.toString(res));if(compare(res, 0, tmp, 0)<0){res = tmp;}}return res;}
}