[leetcode]315. Count of Smaller Numbers After Self

问题描述

You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].

Example:

Given nums = [5, 2, 6, 1]

To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.
Return the array [2, 1, 1, 0].

最开始由编程之美的光影切割问题中求逆序対找到这道题,网上有很多非常巧妙的解法,自己实现了两种。

方法一:改写归并

    private static void reverse(int[] nums, int[] smaller, int[] pos, int start, int end){
        if(start>=end){
            return ;
        }
        int mid=start+(end-start)/2;
        reverse(nums, smaller, pos, start, mid);
        reverse(nums, smaller, pos, mid+1, end);
        int[] temp=new int[end-start+1];
        int i=start, j=mid+1, k=0;
        int count=0;
        while(i<=mid||j<=end){
            if(i>mid){
                temp[k++]=pos[j++];
            }else if(j>end){
                smaller[pos[i]]+=count;
                temp[k++]=pos[i++];
            }else if(nums[pos[i]]<=nums[pos[j]]){
                smaller[pos[i]]+=count;
                temp[k++]=pos[i++];
            }else{
                count++;
                temp[k++]=pos[j++];
            }

        }       
        for(i=0; ipos[start+i]=temp[i];
        }       
    }

    public static List countSmaller(int[] nums) {
        List re=new ArrayList();
        if(nums==null||nums.length==0){
            return re;
        }
        int[] smaller=new int[nums.length];
        int[] pos=new int[nums.length];
        for(int i=0; i<pos.length; i++){
            pos[i]=i;
        }
        reverse(nums, smaller, pos, 0, nums.length-1);
        for(int i=0; ilength; i++){
            re.add(smaller[i]);
        }
        return re;
    }

因为要保存原来的顺序,所以没有直接改写数组,而是用一个pos数组来存储对应的位置。参考博客http://blog.csdn.net/jmspan/article/details/51219203

方法二:树状数组

学习了树状数组和线段树这两种数据结构。下面是用树状数组实现的代码

    private static int[] tree;

    //方法1、通过二分查找 
    public static List countSmaller(int[] nums) {
        int[] temp=nums.clone();
        Arrays.sort(temp);
        for(int i=0; inew int[nums.length];
        Integer[] re=new Integer[nums.length];
        for(int i=nums.length-1; i>=0; i--){
            re[i]=sumRange(nums[i]);
            add(nums[i]+1, 1);
        }
        return Arrays.asList(re);       
    }



    private static void add(int index, int val){
        while(indexindex]+=val;
            index+=lowBit(index);
        }
    }

    private static int sumRange(int index){
        int re=0;
        while(index>0){
            re+=tree[index];
            index-=lowBit(index);
        }
        return re;
    }

    private static int lowBit(int index){
        return index&(-index);
    }

参考博客http://www.cnblogs.com/bigchencheng/p/5151382.html

当然树状数组还有改进的空间,上面把原数组通过二分查找改成0-nums.length的新数组,这样虽然改变了数值,但是相对大小是不变的,因此对答案没有影响。在leetcode讨论区有人给出了如下的改进。

    public static List countSmaller2(int[] nums) {
        pre(nums);
        tree=new int[11000];
        Integer[] re=new Integer[nums.length];
        for(int i=nums.length-1; i>=0; i--){
            re[i]=sumRange(nums[i]);
            add(nums[i]+1, 1);
        }       
        return Arrays.asList(re);       
    }


    private static void pre(int[] nums){
        int min=Integer.MAX_VALUE;
        for(int num: nums){
            min=Math.min(min, num);
        }
        if(min<0){
            min=-min+1;
            for(int i=0; ilength; i++){
                nums[i]+=min;
            }
        }
    }

这种方法虽然空间上存在一定限制,但是避免了用二分求改变原数组,因此在这道题上,时间是最短的。
参考地址https://discuss.leetcode.com/topic/31154/complicated-segmentree-solution-hope-to-find-a-better-one

本题还可以利用二叉搜索树、线段树的方法来解,详细说明参看博客http://www.cnblogs.com/yrbbest/p/5068550.html

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