1365. How Many Numbers Are Smaller Than the Current Number

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

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给定一个数组，对于数组中的每个元素，计算数组中有多少个数小于这个元素。

简单题。因为数组元素的范围很小，在[0, 100]之间，所以直接用hash计算出每个元素出现的次数，然后算出前n项累加和，即可得出小于当前项的元素个数之和，最后查表即可。

完整代码如下：

class Solution {
public:
	vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
		const int MAXN = 101;
		vector<int> ans(nums.size(), 0);
		vector<int> cnt(MAXN, 0);
		for (int i = 0; i < nums.size(); ++i) {
			++cnt[nums[i]];
		}
		for (int i = 1; i < MAXN; ++i) {
			cnt[i] += cnt[i - 1];
		}
		for (int i = 0; i < nums.size(); ++i) {
			if (nums[i] > 0) {
				ans[i] = cnt[nums[i] - 1];
			}
		}
		return ans;
	}
};

本代码提交AC，用时12MS。