How to use multiple stacks to implement a dequeu (using three stacks)
这是 Implement Deque Using Two Stacks 的升华版本， 存一半有效的把时间复杂度最坏情况降低到O(N/2)
这里最好，最坏的情况正好和 Implement Deque Using Two Stacks 相反。
最好的情况: 雨露均沾
最坏的情况：专宠一人
1 /* 2 dequeu can add/remove from two sides 3 stack can add/remove from only one side 4 * */ 5 public class ImplementDequeUsingThreeStacks { 6 private Deque<Integer> stackLeft; 7 private Deque<Integer> stackRight ; 8 private Deque<Integer> stackBuffer; 9 10 public ImplementDequeUsingThreeStacks() { 11 stackLeft = new LinkedList<>(); 12 stackRight = new LinkedList<>(); 13 stackBuffer = new LinkedList<>(); 14 } 15 public void addLeft(int value){ 16 stackLeft.push(value); 17 } 18 public void addRight(int value){ 19 stackRight.push(value); 20 } 21 /* 22 -> <- 23 1 2 | 3 4 24 <- -> 25 this is very interesting: if left is not empty, return left 26 if left is empty, then: 27 1) shuffle the first 1/2 right into the buffer to host, 28 2) and then shift the 2nd half into the left 29 3) pop from the left 30 31 just remember that after each stack shuffle, the order is changed. thats exactly what we need: 3 32 * */ 33 // | 3 4 34 public Integer removeLeft(){ 35 if (!stackLeft.isEmpty()){ 36 return stackLeft.pop(); 37 } 38 //corner case 39 if (isEmpty()){ 40 return null; 41 } else{ 42 //left:[] buffer:[] right:[3,4] 43 int mid = stackRight.size()/2 ; 44 // < since its index based 45 // left:[] buffer: [4] right:[3] 46 for (int i = 0; i <mid ; i++) { 47 stackBuffer.push(stackRight.pop()); 48 } 49 // left:[3] buffer: [4] right:[] 50 while (!stackRight.isEmpty()){ 51 stackLeft.push(stackRight.pop()); 52 } 53 //left:[3] buffer: [] right:[4] 54 while (!stackBuffer.isEmpty()){ 55 stackRight.push(stackBuffer.pop()); 56 } 57 return stackLeft.pop(); 58 } 59 } 60 // 1 2 | 61 public Integer removeRight(){ 62 if (!stackRight.isEmpty()){ 63 return stackRight.pop(); 64 } 65 //corner case 66 if (isEmpty()){ 67 return null; 68 } else{ 69 // left:[1,2] buffer:[] right:[] 70 int mid = stackRight.size()/2 ; 71 // < since its index based 72 // left:[2] buffer: [1] right:[] 73 for (int i = 0; i <mid ; i++) { 74 stackBuffer.push(stackLeft.pop()); 75 } 76 // left:[] buffer: [1] right:[2] 77 while (!stackRight.isEmpty()){ 78 stackRight.push(stackLeft.pop()); 79 } 80 //left:[1] buffer: [] right:[2] 81 while (!stackBuffer.isEmpty()){ 82 stackLeft.push(stackBuffer.pop()); 83 } 84 return stackRight.pop(); 85 } 86 } 87 88 89 public boolean isEmpty(){ 90 return stackLeft.isEmpty() && stackRight.isEmpty() ; 91 } 92 //lazy 93 public int size(){ 94 return stackLeft.size() + stackRight.size() ; 95 } 96 97 public static void main(String[] args) { 98 ImplementDequeUsingThreeStacks deque = new ImplementDequeUsingThreeStacks(); 99 //1 2 | 3 4 100 deque.addLeft(2); 101 deque.addRight(3); 102 deque.addRight(4); 103 deque.addLeft(1); 104 System.out.println(deque.removeLeft());//1 105 System.out.println(deque.removeRight());//4 106 System.out.println(deque.removeRight());//3 107 System.out.println(deque.size()); //1 108 System.out.println(deque.removeLeft()); //2 109 System.out.println(deque.size()); //0 110 } 111 }
