leetcode

class Solution {
    public int strStr(String haystack, String needle) {
        int i = 0, j = 0;
        int a = haystack.length();
        int b = needle.length();
        if(a<b) return -1;
        else if(b==0 || haystack.equals(needle)) return 0;
        else {
            while(i < a && j < b) {
                if(needle.charAt(j) != haystack.charAt(i)) {
                    i=i-j+1;
                    j=0;
                }
                else {
                    i++;
                    j++;
                }
            }
            if(j==b) return i-j;
            else return -1;
        }
    }
}
双指针
class Solution {
    public int strStr(String haystack, String needle) {
                if(haystack.length() == 0&&needle.length() != 0) return -1;
                int l1=haystack.length();
                int l2=needle.length();
                boolean flag;
                for(int i = 0 ; i <= l1-l2; i++){
                    flag = true;
                    for(int j = 0 ; j < needle.length() ; j ++) {
                        if(haystack.charAt(i+j) != needle.charAt(j)) {
                            flag = false;
                            break;
                        }
                    }
                    if(flag) return i;
                }
                return -1;
            }
    }
简洁
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
/*
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
            if(root == null) return res;
            //定义一个队列q
            Queue<TreeNode> q = new LinkedList<>();
            q.offer(root);//把根加入队列
            while(!q.isEmpty()) {
                int sz = q.size();
                ArrayList<Integer> list = new ArrayList<>();
                for(int i=0; i<sz; i++) { //队列大小
                    TreeNode node = q.poll(); //pop
                    list.add(node.val); //push
                    if(node.left != null) q.offer(node.left);
                    if(node.right != null) q.offer(node.right);
                }
                res.add(list);
            }
            return res;
    }
}*/
class Solution {
     /*
      * 队列:Queue<TreeNode> queue=new LinkedList<>();
      * 入队:offer
      * 出队:poll
      * 得到队头:peek
      * 大小:size
      * 空:isEmpty
      */
      public List<List<Integer>> levelOrderBottom(TreeNode root) {
        // write your code here
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> list = new ArrayList<>();
        if(root==null){
            return res;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        TreeNode curLast = root;
        TreeNode nextLast = root;
        queue.offer(root);
        while (!queue.isEmpty()){
            TreeNode node = queue.poll();
            list.add(node.val);
            if(node.left!=null){
                queue.offer(node.left);
                nextLast = node.left;
            }
            if(node.right!=null){
                queue.offer(node.right);
                nextLast = node.right;
            }
            if(node==curLast){
                curLast = nextLast;
                res.add(new ArrayList<>(list));
                list = new ArrayList<>();
            }
        }
        Collections.reverse(res);
        return res;
    }
}
107
原文地址:https://www.cnblogs.com/Roni-i/p/9746350.html