[LeetCode] 315. Count of Smaller Numbers After Self (Hard)
315. Count of Smaller Numbers After Self
class Solution {
public:
vector<int> countSmaller(vector<int>& nums) {
int n = nums.size();
vector<int> v(n);
for (int i = n - 1; i >= 0; --i) {
int val = nums[i];
int L = i + 1, R = n - 1;
while (L <= R) {
int M = L + (R - L) / 2;
if (nums[M] >= val) {
L = M + 1;
} else {
R = M - 1;
}
}
for (int j = i; j < R; ++j) {
nums[j] = nums[j + 1];
}
nums[R] = val;
v[i] = n - R - 1;
}
return v;
}
};
// 1320 ms
算法思想: 从右端折半插入. 对于每个数字来说, 数组长度 - 插入后的位置 - 1
就是该数字的count
.
时间复杂度: O(n^2)
.
空间复杂度: O(1)
.
还有更快的方法. 以后更新.