[LeetCode] 1365. How Many Numbers Are Smaller Than the Current Number

Description

Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i].

Return the answer in an array.

Example 1:

Input: nums = [8,1,2,2,3]
Output: [4,0,1,1,3]
Explanation:
For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3).
For nums[1]=1 does not exist any smaller number than it.
For nums[2]=2 there exist one smaller number than it (1).
For nums[3]=2 there exist one smaller number than it (1).
For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).

Example 2:

Input: nums = [6,5,4,8]
Output: [2,1,0,3]

Example 3:

Input: nums = [7,7,7,7]
Output: [0,0,0,0]

Constraints:

• 2 <= nums.length <= 500
• 0 <= nums[i] <= 100

Analyse

输入一个数组nums，返回一个数组result，数组中的result[i]代表nums中比nums[i]小的元素个数

方法一

先对nums进行排序，排序后的下标就是比nums[i]小的元素个数，存到map里，但如果数组有重复的元素就会出错，所以禁止对map中的数据更新

vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
    vector<int> result(nums); // 备份nums

    sort(nums.begin(), nums.end());

    map<int, int> rankMap;

    int len = nums.size();
    for (int i = 0; i < len; i++) {
        if (rankMap.find(nums[i]) == rankMap.end()) { // 仅当首次赋值时允许
            rankMap[nums[i]] = i;
        }
    }

    for (int j = 0; j < result.size(); j++) {
        result[j] = rankMap[result[j]];
    }

    return result;
}

方法二

LeetCode上还有以一种做法，考虑到0 <= nums[i] <= 100，可以用一个长度为101的数组存储 0-100出现的次数，比nums[i]小的元素的出现次数就是 nums[0] + nums[1] + ...... + nums[i-1]

vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
    vector<int> bitmap(101, 0); // 记录0-100出现次数
    vector<int> result(nums.size(), 0);

    for(int i : nums) {
        ++bitmap[i];
    }

    for(int i = 1; i < 101; i++) { // bitmap[i]现在为比i小的元素出现次数
        bitmap[i] += bitmap[i-1];
    }

    for(int i = 0; i < nums.size(); ++i) {
        result[i] = (nums[i] == 0) ? 0 : bitmap[nums[i] - 1];
    }

    return result;
}