leetcode Count of Smaller Numbers After Self
2015-12-12 22:58
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You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].
Example:
Given nums = [5, 2, 6, 1]
To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.
Return the array [2, 1, 1, 0].
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解题思路:水题一枚,从后向前处理,利用树状数组统计即可。其实应该先将数组中的数据离散化,之后用树状统计即可,无奈数据太水,随便搞搞就过了。
Example:
Given nums = [5, 2, 6, 1]
To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.
Return the array [2, 1, 1, 0].
Subscribe to see which companies asked this question
解题思路:水题一枚,从后向前处理,利用树状数组统计即可。其实应该先将数组中的数据离散化,之后用树状统计即可,无奈数据太水,随便搞搞就过了。
class Solution { public: int tree[1000010]; void init() { memset(tree, 0, sizeof(tree)); } int lowbit(int x) { return x&(-x); } void add(int x, int c) { for(int i = x; i <= 1000000; i += lowbit(i)) { tree[i] += c; } return ; } int getsum(int x) { int ans = 0; for(int i = x; i > 0; i -= lowbit(i)) { ans += tree[i]; } return ans; } vector<int> countSmaller(vector<int>& nums) { vector<int> ans; init(); int size = nums.size(); if(size == 0) return ans; ans.push_back(0); add(nums[size-1]+500000, 1); for(int i = size - 2; i >= 0; --i) { ans.push_back(getsum(nums[i]+500000-1)); add(nums[i]+500000, 1); } reverse(ans.begin(), ans.end()); return ans; } };
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