zjtd-lc-2

简单3星

155. 最小栈

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155. 最小栈

class MinStack {

   int[] data;
        int curLength = 0;
        int curMinLength = 0;
        int[] minStack;

        /**
         * initialize your data structure here.
         */
        public MinStack() {
            data = new int[10];
            minStack = new int[10];
        }

        public int[] grow(int minLength, int[] data) {
            int oldLength = data.length;
            int newLength = oldLength * 2;
            if (minLength > newLength) {
                newLength = minLength;
            }
            // 越界，注意新的长度超过data的长度会越界的
//            System.arraycopy(data, 0, data, 0, newLength);
            // 返回一个新的数组，如果长度超过原来的长度，则补充0
            return Arrays.copyOf(data, newLength);
        }

        public void push(int x) {
            if (curLength + 1 > data.length) {
                data = grow(curLength, data);
            }
            if (curMinLength + 1 > minStack.length) {
                minStack = grow(curMinLength, minStack);
            }
            data[curLength++] = x;
            // 必须带=如果不带，会出现重复的数据，在获取最小值时越界
            if (curMinLength == 0 || minStack[curMinLength - 1] >= x) {
                minStack[curMinLength++] = x;
            }
        }

        public void pop() {
            if (data[curLength - 1] == minStack[curMinLength - 1]) {
                curMinLength--;
            }
            curLength--;
        }

        public int top() {
            return data[curLength - 1];
        }

        public int getMin() {
            return minStack[curMinLength - 1];
        }
}

226. 翻转二叉树

public TreeNode invertTree(TreeNode root) {
    if (root == null) {
        return null;
    }
    TreeNode right = invertTree(root.right);
    TreeNode left = invertTree(root.left);
    root.left = right;
    root.right = left;
    return root;
}
public TreeNode invertTree(TreeNode root) {
    if (root == null) return null;
    Queue queue = new LinkedList();
    queue.add(root);
    while (!queue.isEmpty()) {
        TreeNode current = queue.poll();
        TreeNode temp = current.left;
        current.left = current.right;
        current.right = temp;
        if (current.left != null) queue.add(current.left);
        if (current.right != null) queue.add(current.right);
    }
    return root;
}

234. 回文链表

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69. x 的平方根

class Solution {
  public int mySqrt(int x) {
    if (x < 2) return x;

    long num;
    int pivot, left = 2, right = x / 2;
    while (left <= right) {
      pivot = left + (right - left) / 2;
      num = (long)pivot * pivot;
      if (num > x) right = pivot - 1;
      else if (num < x) left = pivot + 1;
      else return pivot;
    }

    return right;
  }
}

415. 字符串相加

class Solution {
    public String addStrings(String num1, String num2) {
        StringBuilder res = new StringBuilder("");
        int i = num1.length() - 1, j = num2.length() - 1, carry = 0;
        while(i >= 0 || j >= 0){
            int n1 = i >= 0 ? num1.charAt(i) - '0' : 0;
            int n2 = j >= 0 ? num2.charAt(j) - '0' : 0;
            int tmp = n1 + n2 + carry;
            carry = tmp / 10;
            res.append(tmp % 10);
            i--; j--;
        }
        if(carry == 1) res.append(1);
        return res.reverse().toString();
    }
}

83. 删除排序链表中的重复元素

public ListNode deleteDuplicates(ListNode head) {
    ListNode current = head;
    while (current != null && current.next != null) {
        if (current.next.val == current.val) {
            current.next = current.next.next;
        } else {
            current = current.next;
        }
    }
    return head;
}

28. 实现 strStr()

class Solution {
  public int strStr(String haystack, String needle) {
    int L = needle.length(), n = haystack.length();
    if (L == 0) return 0;

    int pn = 0;
    while (pn < n - L + 1) {
      // find the position of the first needle character
      // in the haystack string
      while (pn < n - L + 1 && haystack.charAt(pn) != needle.charAt(0)) ++pn;

      // compute the max match string
      int currLen = 0, pL = 0;
      while (pL < L && pn < n && haystack.charAt(pn) == needle.charAt(pL)) {
        ++pn;
        ++pL;
        ++currLen;
      }

      // if the whole needle string is found,
      // return its start position
      if (currLen == L) return pn - L;

      // otherwise, backtrack
      pn = pn - currLen + 1;
    }
    return -1;
  }
}

KMP看解题

13. 罗马数字转整数

class Solution {
    public int romanToInt(String s) {
        Map map = new HashMap<>();
        map.put("I", 1);
        map.put("IV", 4);
        map.put("V", 5);
        map.put("IX", 9);
        map.put("X", 10);
        map.put("XL", 40);
        map.put("L", 50);
        map.put("XC", 90);
        map.put("C", 100);
        map.put("CD", 400);
        map.put("D", 500);
        map.put("CM", 900);
        map.put("M", 1000);
        
        int ans = 0;
        for(int i = 0;i < s.length();) {
            if(i + 1 < s.length() && map.containsKey(s.substring(i, i+2))) {
                ans += map.get(s.substring(i, i+2));
                i += 2;
            } else {
                ans += map.get(s.substring(i, i+1));
                i ++;
            }
        }
        return ans;
    }
}

167. 两数之和 II - 输入有序数组

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int low = 0;
        int high = numbers.length - 1;
        while(low < high){
            int sum = numbers[low] + numbers[high];
            if(sum == target){
                return new int[]{low + 1, high + 1};
            }else if(sum > target){
                high--;
            }else{
                low++;
            }
        }
        return new int[]{-1,-1};
    }
}

203. 移除链表元素

class Solution {
  public ListNode removeElements(ListNode head, int val) {
    ListNode sentinel = new ListNode(0);
    sentinel.next = head;

    ListNode prev = sentinel, curr = head;
    while (curr != null) {
      if (curr.val == val) prev.next = curr.next;
      else prev = curr;
      curr = curr.next;
    }
    return sentinel.next;
  }
}

35. 搜索插入位置

class Solution {
    public int searchInsert(int[] nums, int target) {
        
        if(nums.length==0){
            return 0;
        }
        int left = 0;
        int right = nums.length;
        while(left nums[mid]){
                left = mid+1;
            }else{
                right = mid;
            }
        }
        return right;

    }
}

136. 只出现一次的数字

class Solution {
  public int singleNumber(int[] nums) {
    int a = 0;
    for (int i : nums) {
      a ^= i;
    }
    return a;
  }
}

108. 将有序数组转换为二叉搜索树

class Solution {
  int[] nums;

  public TreeNode helper(int left, int right) {
    if (left > right) return null;

    // always choose left middle node as a root
    int p = (left + right) / 2;

    // inorder traversal: left -> node -> right
    TreeNode root = new TreeNode(nums[p]);
    root.left = helper(left, p - 1);
    root.right = helper(p + 1, right);
    return root;
  }

  public TreeNode sortedArrayToBST(int[] nums) {
    this.nums = nums;
    return helper(0, nums.length - 1);
  }
}

225. 用队列实现栈

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面试题03. 数组中重复的数字

class Solution {
    public int findRepeatNumber(int[] nums) {
        Set set = new HashSet();
        int repeat = -1;
        for (int num : nums) {
            if (!set.add(num)) {
                repeat = num;
                break;
            }
        }
        return repeat;
    }
}

118. 杨辉三角

class Solution {
    public List> generate(int numRows) {
        List> triangle = new ArrayList>();

        // First base case; if user requests zero rows, they get zero rows.
        if (numRows == 0) {
            return triangle;
        }

        // Second base case; first row is always [1].
        triangle.add(new ArrayList<>());
        triangle.get(0).add(1);

        for (int rowNum = 1; rowNum < numRows; rowNum++) {
            List row = new ArrayList<>();
            List prevRow = triangle.get(rowNum-1);

            // The first row element is always 1.
            row.add(1);

            // Each triangle element (other than the first and last of each row)
            // is equal to the sum of the elements above-and-to-the-left and
            // above-and-to-the-right.
            for (int j = 1; j < rowNum; j++) {
                row.add(prevRow.get(j-1) + prevRow.get(j));
            }

            // The last row element is always 1.
            row.add(1);

            triangle.add(row);
        }

        return triangle;
    }
}

38. 外观数列

class Solution {
    public String countAndSay(int n) {
        String dp= "1" ;
        StringBuilder sb = new StringBuilder() ;
        for(int i = 1 ; i < n ; i++){
            int count = 1 ;
            sb = new StringBuilder() ;
            char[] ss = dp.toCharArray() ;
            int len = ss.length ;
            for(int j = 0 ; j < len - 1 ; j++){
                if(ss[j] == ss[j+1]){
                    count++ ;
                    continue ;
                }
                sb.append(String.valueOf(count)).append(ss[j]) ;
                count = 1 ;   
            }
            sb.append(String.valueOf(count)).append(ss[len-1]) ;
            dp = sb.toString() ;
        }
        return dp ;
    }
}

115. 不同的子序列

https://leetcode-cn.com/problems/distinct-subsequences/solution/yi-wei-shu-zu-dpxiang-xi-jie-shi-by-dijks/

面试题62. 圆圈中最后剩下的数字

约瑟夫环
https://leetcode-cn.com/problems/yuan-quan-zhong-zui-hou-sheng-xia-de-shu-zi-lcof/solution/huan-ge-jiao-du-ju-li-jie-jue-yue-se-fu-huan-by-as/
f(N,M)=(f(N−1,M)+M)%n

232. 用栈实现队列

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67. 二进制求和

class Solution {
    public String addBinary(String a, String b) {
        int ca = 0;
        StringBuilder ans = new StringBuilder();
        for(int i = a.length()-1,j=b.length()-1;i>=0||j>=0;i--,j--){
            int sum = ca + (i>=0 ? a.charAt(i) - '0' : 0) + (j>=0 ? b.charAt(j) - '0' : 0);
            ca = sum / 2;
            ans.insert(0, sum % 2);
        }
        if(ca == 1) {
            ans.insert(0,1);
        }
        return ans.toString();
    }
}

557. 反转字符串中的单词 III

public class Solution {
    public String reverseWords(String s) {
        String words[] = split(s);
        StringBuilder res=new StringBuilder();
        for (String word: words)
            res.append(reverse(word) + " ");
        return res.toString().trim();
    }
    public String[] split(String s) {
        ArrayList < String > words = new ArrayList < > ();
        StringBuilder word = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == ' ') {
                words.add(word.toString());
                word = new StringBuilder();
            } else
                word.append( s.charAt(i));
        }
        words.add(word.toString());
        return words.toArray(new String[words.size()]);
    }
    public String reverse(String s) {
      StringBuilder res=new StringBuilder();
        for (int i = 0; i < s.length(); i++)
            res.insert(0,s.charAt(i));
        return res.toString();
    }
}

557. 反转字符串中的单词 III

https://leetcode-cn.com/problems/intersection-of-two-arrays/solution/duo-chong-jie-fa-jie-jue-349-liang-ge-shu-zu-de-ji/

66. 加一

class Solution {
    public int[] plusOne(int[] digits) {
        int length = digits.length;
        for(int i = length-1;i>=0 ;i--){
            digits[i]++;
            digits[i] %=10;
            if(digits[i]!=0){
                return digits;
            }
        }
        digits = new int[length+1];
        digits[0] = 1;
        return digits;
    }
}