算法练习LeetCode初级算法之设计问题

• 打乱数组

不断的让第一个与后面随机选择的数交换

class Solution {

    private int[] nums;

    private int[] initnums;

public Solution(int[] nums) {

    this.nums=nums;

    this.initnums=Arrays.copyOf(nums, nums.length);//这里必须要复制，要指明对象

}

 

/** Resets the array to its original configuration and return it. */

public int[] reset() {

    return initnums;//这里返回上面的复制

}

 

/** Returns a random shuffling of the array. */

public int[] shuffle() {

    Random random=new Random();

    for (int i = 0; i <nums.length/2; i++) {

            swap(nums, 0, random.nextInt(nums.length));

        }

    return nums;

}

private void swap(int[] nums,int i,int j) {

        int temp=nums[i];

        nums[i]=nums[j];

        nums[j]=temp;

    }

}

• 最小栈

• 自己写的有点慢，勉强通过

  class MinStack {

  /** initialize your data structure here. */

      Stack<Integer> stack;

  public MinStack() {

      stack=new Stack<>();

  }

  public void push(int x) {

  stack.push(x);

  }

  public void pop() {

      if (!stack.isEmpty()) {

              stack.pop();

          }

  }

  public int top() {

       return stack.peek();//这里和我之前理解的不太一样，之前是取栈顶顺便就删除了

  }

  public int getMin() {

      Stack<Integer> stack2=new Stack<>();

      stack2.addAll(0, stack);

  Collections.sort(stack2);

  return stack2.get(0);

  }

  }

• 双栈法挺快

  class MinStack {

  /** initialize your data structure here. */

      Stack<Integer> stack,minStack;

  public MinStack() {

      stack=new Stack<>();

      minStack=new Stack<>();

  }

  public void push(int x) {

  stack.push(x);

  if (minStack.isEmpty()) {

              minStack.push(x);

          }else if (x<=minStack.peek()) {//此处若没有等号

              minStack.push(x);

          }

  }

  public void pop() {

      if (stack.peek().equals(minStack.peek())) {//这里可能会提示栈为空的异常

          stack.pop();

          minStack.pop();

          }else {

              stack.pop();

          }

  }

  public int top() {

       return stack.peek();

  }

  public int getMin() {

  return minStack.peek();

  }

  }