leetcode [Min Stack]

public class MinStack {    /** initialize your data structure here. */private TreeMap<Integer, Integer> min;//用TreeMap来对min的序列有序排列private LinkedList<Integer> stack;    public MinStack() {    min = new TreeMap<>();    stack = new LinkedList<>();    }        public void push(int x) {        stack.add(x);        if(min.containsKey(x)){        min.put(x, min.get(x) + 1);//相同的key会覆盖,map.get(key)返回的是key的value        }        else{        min.put(x, 1);        }    }        public void pop() {        int pop = stack.removeLast();        //min.remove(pop);//假如插入时被去重了怎么办，所以用TreeMap的value来记录出现次数        if(min.get(pop) == 1){        min.remove(pop);        }        else{        min.put(pop, min.get(pop) - 1);        }    }        public int top() {        return stack.getLast();    }        public int getMin() {        return min.firstKey();//因为TreeMap是有序的，所以TreeMap的第一个元素就是最小值    }}/** * Your MinStack object will be instantiated and called as such: * MinStack obj = new MinStack(); * obj.push(x); * obj.pop(); * int param_3 = obj.top(); * int param_4 = obj.getMin(); */

解法二：

public class MinStack {    /** initialize your data structure here. */private int min;//比起解法一// only push the old minimum value when the current minimum value [changes] after pushing the new value xprivate LinkedList<Integer> stack;    public MinStack() {    min = Integer.MAX_VALUE;//让min初始化为最大后面才好更新    stack = new LinkedList<>();    }        public void push(int x) {    // only push the old minimum value when the current minimum value [changes] after pushing the new value x        if(x <= min){            stack.add(min);//压入旧的min值，因为栈的最底部是Integer.MAX_VALUE，若要继续完善，所以计数时判断isEmpty基数是1        min = x;//压栈过程中更新min的值        }        stack.add(x);//正常压入元素，先压入旧的min，后压入元素    }        public void pop() {        int pop = stack.removeLast();        if(pop == min){//说明弹出的元素是min        min = stack.removeLast();//找一个旧的min来给min，因为压入时【先压入旧的min，后压入元素】        }    }        public int top() {        return stack.getLast();    }        public int getMin() {        return min;    }}/** * Your MinStack object will be instantiated and called as such: * MinStack obj = new MinStack(); * obj.push(x); * obj.pop(); * int param_3 = obj.top(); * int param_4 = obj.getMin(); */

解法三：

public class MinStack {    /** initialize your data structure here. */private long min;//因为这里存在差值，所以用long来存，防止溢出private LinkedList<Long> stack;    public MinStack() {    min = Long.MAX_VALUE;    stack = new LinkedList<>();    }        public void push(int x) {    //解法三：压入栈的第一个元素是0(因为此时min - x =0)，后面压入栈的是min与元素的差值来代表元素，而min是实际的元素，注意min是实时更新的    if(stack.isEmpty()){    stack.add(0L);    min = x;    }    else{    stack.add(x - min);//以旧的min来作差，因为后面的min会被弹出    if(x <= min){//if(x < min)也行，重复的不影响更新    min = x;//更新min的值    }    }    }        public void pop() {    long p = stack.removeLast();    //if p ==0, it means there are duplicates of min value. after pop, the min value did not change.    if(p < 0){//说明x - min = p < 0，在压入时min更新了，那么弹出时min需要更新到上一层的    min = min - p;    }    }        public int top() {        long t = stack.getLast();        if(t > 0){//说明x - min = t > 0，那么是旧的min        return (int)(t + min);        }        else{//说明x - min = t <= 0，那么min是被更新成了x，min是新的了，直接return min，不能再return (t + min)        return (int)min;        }    }        public int getMin() {        return (int)min;    }}/** * Your MinStack object will be instantiated and called as such: * MinStack obj = new MinStack(); * obj.push(x); * obj.pop(); * int param_3 = obj.top(); * int param_4 = obj.getMin(); */