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LeetCode-315 Count of Smaller Numbers After Self

Description

Given an integer array nums, return an integer array counts where \(counts[i]\) is the number of smaller elements to the right of \(nums[i]\).

Constraints

  • \(1 \le nums.length \le 10^5\)
  • \(-10^4 \le nums[i] \le 10^4\)

Solution

模板题,回顾一下也好

Code

class Solution {
public:
    vector<int> countSmaller(vector<int>& nums) {
        constexpr size_t N = 5e4;
        constexpr int shift = 1 + 1e4;
        int ft[N];
        memset(ft, 0, sizeof ft);
        vector<int> ans;
        for(auto iter = nums.rbegin(); iter != nums.rend(); iter++) {
            int v = *iter + shift;
            int sum = 0;
            for(int u = v-1; u; u -= u&-u) sum+=ft[u];
            ans.push_back(sum);
            for(; v < N; v += v&-v) ft[v]++;
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};

LeetCode-315 Count of Smaller Numbers After Self

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