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Update 315. Count of Smaller Numbers After Self.md
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@ -23,11 +23,14 @@
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此题还可以用线段树/树状数组做，我们知道线段树和树状数组可以求前缀和，而这题可以转换成求前缀和。具体转换过程如下：
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1. 我们先遍历一遍数组，确定数组中元素的最小值`MIN`和`MAX`，然后想象有一个大小为`MAX - MIN`的全0数组`arr`，在这个数组上构建线段树/树状数组；
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2. 然后从后往前遍历数组nums，将`arr[nums[i]]++`，更新线段树/树状数组，这样`arr`在区间`(nums[i], MAX]`的元素和即为`nums[i]`右侧它小的元素个数。
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1. 我们先遍历一遍数组，确定数组中元素的最小值`MIN`和`MAX`，然后想象有一个大小为`MAX - MIN`的全0数组`arr`，`arr[i]=j`表示`i+MIN`出现了j次。然后在这个数组上构建线段树/树状数组；
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2. 然后从后往前遍历数组nums，将`arr[nums[i]-MIN]++`，表示出现次数加1，更新线段树/树状数组，这样就可很方便求得右侧比它小的元素个数。
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时间复杂度O(nlogN)，空间复杂度O(N)，其中`N = MAX - MIN`；
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关于线段树和树状数组可参考我的博客[Range Sum Query - Mutable (区间查询)](https://tangshusen.me/2019/11/17/range-sum-query-mutable/)
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# C++
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## 思路一
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``` C++
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@ -119,6 +122,71 @@ public:
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}
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};
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```
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## 思路三、线段树
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``` C++
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class SegTree{
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private:
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int n;
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vector<int>tree; // 用一个长为2n的数组来表示树
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public:
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SegTree(vector<int>& nums) {
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/* 对数组nums建线段树, 方便求sum(nums[i,...,j])以及更新nums[i]*/
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n = nums.size();
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tree = vector<int>(n*2, 0);
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for(int i = 0; i < n; i++) // 建树
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update(i, nums[i]);
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}
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void update(int i, int diff) {
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/*更新操作: 将nums[i]的值加上diff*/
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i += n; // 转换为线段树下标
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while(i > 0){
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tree[i] += diff;
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i >>= 1;
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}
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}
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int sumRange(int i, int j) {
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/**求nums[i,...,j]的和*/
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if(i > j) return 0;
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i += n; j += n; // 转换为线段树下标
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int res = 0;
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for(; i <= j; i >>= 1, j >>=1){
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if(i & 1) res += tree[i++]; // 是右孩子
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if(!(j & 1)) res += tree[j--];
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}
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return res;
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}
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};
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class Solution {
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public:
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vector<int> countSmaller(vector<int>& nums) {
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int n = nums.size();
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vector<int>res(n, 0);
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if(!n) return res;
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int min_num = nums[0], max_num = nums[0];
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for(int &num: nums){
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min_num = min(min_num, num);
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max_num = max(max_num, num);
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}
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// nums_for_SegTree[i] = j 表示数字i出现了j次, 初始全0次
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vector<int>nums_for_SegTree(max_num - min_num + 1, 0);
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SegTree st = SegTree(nums_for_SegTree);
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for(int i = n - 1; i >= 0; i--){
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res[i] = st.sumRange(0, nums[i] - min_num - 1);
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st.update(nums[i] - min_num, 1); // 出现次数+1, 更新树
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}
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return res;
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}
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};
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```
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## 思路三、树状数组
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[来源](https://leetcode-cn.com/problems/count-of-smaller-numbers-after-self/solution/c-shu-zhuang-shu-zu-by-mryx/)
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@ -164,108 +232,3 @@ public:
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}
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};
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```
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## 思路三、线段树
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[来源](https://leetcode-cn.com/problems/count-of-smaller-numbers-after-self/solution/c-xian-duan-shu-jie-fa-by-dufre/)
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``` C++
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struct SegmentTreeNode{
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int start;
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int end;
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int count;
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SegmentTreeNode* left;
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SegmentTreeNode* right;
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SegmentTreeNode(int _start, int _end):start(_start),end(_end) {
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count = 0;
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left = NULL;
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right = NULL;
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}
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};
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class Solution {
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public:
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SegmentTreeNode* build(int start, int end){
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if (start > end)
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return NULL;
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SegmentTreeNode* root = new SegmentTreeNode(start, end);
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if (start == end){
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root->count = 0;
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}else{
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int mid = start + (end - start)/2;
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root->left = build(start, mid);
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root->right = build(mid+1, end);
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}
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return root;
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}
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int count(SegmentTreeNode* root, int start, int end){
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if (root == NULL || start>end)
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return 0;
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if (start==root->start && end==root->end){
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return root->count;
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}
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int mid = root->start + (root->end - root->start)/2;
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int leftcount = 0, rightcount = 0;
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if (start <= mid){
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if (mid < end)
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leftcount = count(root->left, start, mid);
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else
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leftcount = count(root->left, start, end);
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}
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if (mid < end){
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if (start <= mid)
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rightcount = count(root->right, mid+1, end);
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else
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rightcount = count(root->right, start, end);
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}
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return (leftcount + rightcount);
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}
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void insert(SegmentTreeNode* root, int index, int val){
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if (root->start==index && root->end==index){
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root->count += val;
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return;
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}
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int mid = root->start + (root->end - root->start)/2;
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if (index>=root->start && index<=mid){
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insert(root->left, index, val);
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}
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if (index>mid && index<=root->end){
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insert(root->right, index, val);
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}
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root->count = root->left->count + root->right->count;
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}
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vector<int> countSmaller(vector<int>& nums) {
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vector<int> res;
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if (nums.empty())
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return res;
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res.resize(nums.size());
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int start = nums[0];
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int end = nums[0];
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for (int i=1; i<nums.size(); i++){
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start = min(start, nums[i]);
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end = max(end, nums[i]);
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}
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SegmentTreeNode* root = build(start, end);
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for (int i=nums.size()-1; i>=0; i--){
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res[i] = count(root, start, nums[i]-1);
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insert(root, nums[i], 1);
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}
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return res;
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}
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};
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```
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