LeetCode 2300. 咒语和药水的成功对数

2300. 咒语和药水的成功对数

 

【二分】

class Solution {

    // 二分 4:51 5

    int bs(int[] arr, double target){
        int left = 0, right = arr.length - 1;
        while(left <= right){
            int mid = (left + right) >>> 1;
            if(arr[mid] >= target) right = mid - 1;
            else left = mid + 1;
        }
        return left;
    }

    public int[] successfulPairs(int[] ss, int[] ps, long k) {
        Arrays.sort(ps);
        int n = ss.length, m = ps.length;
        int[] ans = new int[n];
        for(var i = 0; i < n; i++){
            double t = k * 1.0 / ss[i];
            ans[i] = m - bs(ps, t);
        }
        return ans;
    }
}

【手写快排+二分】

唉，不太想写了说实话

class Solution {

    // 7:11

    int[] ps, ss;
    int m, n;

    void qs(int left, int right){
        if(left >= right) return;
        int l = left, r = right, flag = ps[left];
        while(l < r){
            while(l < r && ps[r] > flag) r--;
            ps[l] = ps[r];
            while(l < r && ps[l] <= flag) l++;
            ps[r] = ps[l];
        } 
        ps[l] = flag;
        qs(left, l - 1);
        qs(l + 1, right);
    }

    int bs(long target){
        int left = 0, right = ps.length - 1;
        while(left <= right){
            int mid = (left + right) >>> 1;
            if(ps[mid] >= target) right = mid - 1;
            else left = mid + 1;
        }
        return left;
    }

    public int[] successfulPairs(int[] spells, int[] potions, long k) {
        ss = spells; ps = potions; n = ps.length; m = ss.length;
        qs(0, n - 1);
        int[] ans = new int[m];
        for(int i = 0; i < m; i++){
            ans[i] = n - bs((k + ss[i] - 1) / ss[i]);
        }
        return ans;
    }
}

 

果然TLE了，因为触发到快排最差的情况了（基本有序的情况）快排退化到O(n^2)了