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Given an array and a value, remove all instances of that value in place and return the new length. Do not allocate extra space for another array, you must do this in place with constant memory. The order of elements can be changed. It doesn’t matter what you leave beyond the new length. This problem is similar to another leetcode problem: Remove Duplicates from Sorted Array Do not allocate extra space for another array, you must do this in place with constant memory. 　　You are given a target value to search. If found in the array return its index, otherwise return -1. Using the idea of binary search, we can reduce the search space based on which range the target value is located.  It is easy to use an intermediate array to solve this problem. Suppose the original array is A, the new array is B, we will copy  Evaluate RPN is the second step to build a calculator. Once we have converted the infix expression into the RPN The basic idea to evaluate reverse polish notation is to use a stack to process the strings. Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). Bonus points if you could solve it both recursively and iteratively. This problem can be easily solved using a recursive algorithm.  We defined a method with the following signature: then we recursively call equal(left.left, right.right) and equal(left.right, right.left).  Window position                Max ---------------               -----         res = Arrays.copyOf(nums, nums.length);         r = new Random();         int right = res.length - 1;         int j;         while(right > 0) {             j = r.nextInt(right + 1);             swap(res, j, right);             right--; Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list. Given word1 = “coding”, word2 = “practice”, return 3. Given word1 = “makes”, word2 = “coding”, return 1.