135. Candy

 follow up：如何不使用extra space解决这道题

https://leetcode.com/problems/candy/solution/

Approach 4: Single Pass Approach with Constant Space
Algorithm



https://leetcode.com/problems/candy/discuss/42770/One-pass-constant-space-Java-solution

利用BST中序遍历是单调递增的性质。维护一个curVal，在中序遍历的时候，若是root.val == curVal，更新curFreq。由于BST保证了你只会在连续几个node的时候碰到相同的val.不可能你刚开始遍历了几个4，而后遍历了几个5，而后又遍历到4.


 给你一个BST, 可是这个BST是有duplicate的，问你在这棵BST中出现频率最高的那个数字是什么，以及频率是多少。 follow up是， 不用extra space

请问楼主，生成迷宫能够用 https://en.wikipedia.org/wiki/Maze_generation_algorithm 上说的 Recursive backtracker 写吗？感受这个方法还挺直接的

我以为这个挺对的！稍微变种一下这个recursive backtracker的思想就好。wiki上的例子wall是不占用一个格子的，因此wiki上的goal是要遍历全部的cell。但咱们这里wall要占用格子，那么咱们能够把全部i, j 为偶数的(i, j) 格子当作cell，其余的格子当作是wall，目标就是把全部的cell格子遍历一遍。这样知足了题目中不能产生2*4thick wall 的条件，由于wall最厚也只能是1.

]]
There are N children standing in a line. Each child is assigned a rating value.
You are giving candies to these children subjected to the following requirements:
* Each child must have at least one candy.
* Children with a higher rating get more candies than their neighbors.
What is the minimum candies you must give?
Example 1:
Input: [1,0,2]
Output: 5
Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.
Example 2:
Input: [1,2,2]
Output: 4
Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
             The third child gets 1 candy because it satisfies the above two conditions.




https://www.youtube.com/watch?v=-XWLumbr4UU


https://leetcode.com/problems/candy/discuss/42774/Very-Simple-Java-Solution-with-detail-explanation



class Solution {
    public int candy(int[] ratings) {
        
        // fill in the array with 1 first , and then iterate from left to right 
        // if next one is bigger , then next one is cur one + 1 , 
        // next one is not bigger, then next one stays the same 
        
        // iterate from right to left, if next one is bigger , next one is max(cur + 1, next one)
        // if next one is not bigger, next one stays the same 
        
        // then iterate thru the whole array and get the result 
        // or we can update the result the second traversal from right to left
        // 1 0 2 
        // 1 1 2 
        int[] array = new int[ratings.length];
        Arrays.fill(array, 1);
        for(int i = 0; i < array.length - 1; i++){  // i < 3 - 1 = 2 , i < 2, i = 0, 1 
            if(ratings[i + 1] > ratings[i]){
                array[i + 1] = array[i] + 1;
                // i + 1 = 1, 2
            }
        }
        // from right to left 
        int res = 0;
        for(int i = array.length - 1; i > 0; i--){
            if(ratings[i - 1] > ratings[i]){
                array[i - 1] = Math.max(array[i - 1], array[i] + 1);
            }
        }
        for(int i = 0; i < array.length; i++){
            res += array[i];
        }
        return res;
    }
}