Du Yao's blog

91. Decode Ways 方法一暴力递归–超时 1234567891011121314151617181920212223242526272829303132333435363738394041public int numDecodings(String s) { if(s == null || s.length() == 0){ return 0; }... 阅读全文...

307. Range Sum Query - Mutable 位运算 运算 含义 x>>1 右移一位，x /= 2 x<<1 左移一位，x *= 2 `x 1` 或，如果x = 2*i,那么`x 1` = 2*i +1 x^1 异或，奇偶互换 x&1 是否为奇数，x%2 线段树segment tree 原型是二叉树，叶子节点是... 阅读全文...

78. Subsets相似问题46. Permutations题解46 Answer 方法一思路利用bitmap所给数组是a[n]，则所有的方法总数是2^n的个数所有的组合形式是0-2^n-1的二进制数 >Number of subsets for {1 , 2 , 3 } = 2^3 . why ?case possible outcomes for the set of su... 阅读全文...

93. Restore IP Addresses Backtracking - 回溯 1234567891011121314151617181920212223242526272829303132public List<String> restoreIpAddresses(String s) { List<String> list = new ArrayLi... 阅读全文...

15. 3SumTwo sum的进阶 思路一利用HashMap存储另外两个数字的和，然后遍历数组类似于Two sum的思路但是3重循环，超时1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859public List<Lis... 阅读全文...

普通素数筛法1234567891011121314151617//bitset 被置1说明是素数public static void count(BitSet set){ //length() -> Returns the "logical size" of this BitSet for(int i = 2; i < set.size(); i++... 阅读全文...

1. Two Sum 12345678910111213public int[] twoSum(int[] nums, int target) { HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); for (int i = 0; i < nums.length; i++)... 阅读全文...

230. Kth Smallest Element in a BST 123456789101112131415161718192021public int kthSmallest(TreeNode root, int k) { int cnt = vis(root.left); if(k == cnt+1){ return root.val; }else i... 阅读全文...

130. Surrounded Regions 思路是图的遍历，类似于岛屿的个数但是边缘的全部不能被该成X因此可以先遍历所有边缘的O，即把本身和与其相连的O全部变为b 然后再顺序遍历，将b变为O，将O变为X 注意bfs遍历时，访问元素可以是入队也可以是出队 入队 如果不设置入队判断，必须是入队访问，即入队是将O变为b因为元素可能会重复入内 出队 出队访问的话，有可能会导致元素重复入队... 阅读全文...

Callable与Runnable相似，只是Runnable无返回值A task that returns a result and may throw an exception. Implementors define a single method with no arguments called call.The Callable interface is similar to Run... 阅读全文...