LC.155. Min Stack(优化，针对整块一样数传入)
如果有特别多重复的数进来的情况，并且是一整块一样的数进来，那是可以被优化的。
1111111 222222 -1-1-1-1 111111 -2-2-2-2-2 33333
如果 进来的数字是交叉的，则优化空间有限: 1212121212
1 public class LC_155_MinStack_II { 2 3 private Deque<Integer> stack ; 4 private Deque<Integer> minStack ; 5 //keep each minStack item's times 6 private Deque<Integer> counter ; 7 8 /** initialize your data structure here. */ 9 public LC_155_MinStack_II() { 10 stack = new LinkedList<>() ; 11 minStack = new LinkedList<>(); 12 counter = new LinkedList<>(); 13 } 14 /* 15 stack -2 -2 -2 0 0 -3 16 min -2 -3 17 counter 5 1 18 * */ 19 public void push(int x) { 20 stack.push(x); 21 if (minStack.isEmpty()){ 22 minStack.push(x); 23 counter.push(1); //initialize new item 24 } else{ 25 if (minStack.peek() < x){ 26 int times = counter.pop(); 27 counter.push(times+1); 28 } else{ 29 minStack.push(x); 30 counter.push(1); 31 } 32 } 33 } 34 /* 35 stack -2 -2 -2 0 0 -3 36 min -2 -3 37 counter 5 1 38 * */ 39 public void pop() { 40 if (stack.isEmpty()) return ; 41 stack.pop(); 42 //handle the min: only pop minStack when the counter's item reach 0 43 int times = counter.pop(); 44 if (times>1){ 45 times--; 46 counter.push(times); 47 } 48 //when counter == 1 or less(shouldnt possible be <0), then directly pop counter 49 else { 50 minStack.pop(); 51 } 52 } 53 //peek 54 public int top() { 55 if (stack.isEmpty()) return -1 ; 56 return stack.peek() ; 57 } 58 59 public int getMin() { 60 if (minStack.isEmpty()) return -1 ; 61 return minStack.peek(); 62 } 63 }
