LeetCode - Fruit Into Baskets
阿新 • • 發佈:2018-11-19
In a row of trees, the i-th tree produces fruit with type tree[i]. You start at any tree of your choice, then repeatedly perform the following steps: Add one piece of fruit from this tree to your baskets. If you cannot, stop. Move to the next tree to the right of the current tree. If there is no tree to the right, stop. Note that youdo not have any choice after the initial choice of starting tree: you must perform step 1, then step 2, then back to step 1, then step 2, and so on until you stop. You have two baskets, and each basket can carry any quantity of fruit, but you want each basket to only carry one type of fruit each. What is the total amount of fruit you can collect withthis procedure? Example 1: Input: [1,2,1] Output: 3 Explanation: We can collect [1,2,1]. Example 2: Input: [0,1,2,2] Output: 3 Explanation: We can collect [1,2,2]. If we started at the first tree, we would only collect [0, 1]. Example 3: Input: [1,2,3,2,2] Output: 4 Explanation: We can collect [2,3,2,2]. If we started at the first tree, we would only collect [1, 2]. Example 4: Input: [3,3,3,1,2,1,1,2,3,3,4] Output: 5 Explanation: We can collect [1,2,1,1,2]. If we started at the first tree or the eighth tree, we would only collect 4 fruits. Note: 1 <= tree.length <= 40000 0 <= tree[i] < tree.length
用 two points 去維護一個 window, 注意一些coner case
class Solution { public int totalFruit(int[] tree) { if(tree == null || tree.length == 0){ return 0; } int len = tree.length; int p = 0; int q = 0; int max = 0; Set<Integer> set = new HashSet<>(); while(p < len){ if(set.size() == 0){ set.add(tree[p]); p++; } else if(set.contains(tree[p])){ p++; } else{ if(set.size() == 1){ set.add(tree[p]); p++; } else{ if(p - q > max){ max = p - q; } set.clear(); int temp = p - 1; int type = tree[temp]; set.add(tree[p - 1]); set.add(tree[p]); while(tree[temp] == type){ temp --; } q = temp+1; } } } if(p - q > max){ max = p - q; } return max; } }