想办个网站怎么做中国关于生态文明建设的网站
1、堆排
2、快排
颜色分类
- 颜色分类
class Solution {
public:void sortColors(vector<int>& nums) {int i = 0, l = -1, r = nums.size();while (i < r){if (nums[i] == 0) swap(nums[++l], nums[i++]);else if (nums[i] == 1) i++;else swap(nums[--r], nums[i]);}}
};
排序数组
- 排序数组
class Solution {
public:vector<int> sortArray(vector<int>& nums) {srand(time(nullptr));qsort(nums, 0, nums.size() - 1);return nums;}void qsort(vector<int>& nums, int l, int r){if (l >= r) return;int key = nums[rand() % (r - l + 1) + l]; // 获取随机基准值int i = l, left = l - 1, right = r + 1;// 三路划分while (i < right){if (nums[i] < key) swap(nums[++left], nums[i++]);else if (nums[i] == key) i++;else swap(nums[--right], nums[i]);}qsort(nums, l, left);qsort(nums, right, r);}
};
数组中的第K个最大元素
- 数组中的第K个最大元素
最容易想到的就是优先级队列。
class Solution {
public:int findKthLargest(vector<int>& nums, int k) {priority_queue<int> pq;for (auto e : nums) pq.push(e);int ret = 0;while (k--){ret = pq.top();pq.pop();}return ret;}
};
快速选择算法
经常出错：
else return qsort(arr, l, left, k - b - c);
class Solution {
public:int findKthLargest(vector<int>& nums, int k) {srand(time(nullptr));return qsort(nums, 0, nums.size() - 1, k);}int qsort(vector<int>& arr, int l, int r, int k){if (l >= r) return arr[l];int key = arr[rand() % (r - l + 1) + l]; // 随机基准值int i = l, left = l - 1, right = r + 1;// 三路划分while (i < right){if (arr[i] < key) swap(arr[++left], arr[i++]);else if (arr[i] == key) i++;else swap(arr[--right], arr[i]);}// 分情况讨论int c = r - right + 1, b = right - left - 1;if (c >= k) return qsort(arr, right, r, k);else if (c + b >= k) return key;else return qsort(arr, l, left, k - b - c);}
};
LCR 159. 库存管理 III
- LCR 159. 库存管理 III
最先想到的还是优先级队列。
class Solution {
public:vector<int> inventoryManagement(vector<int>& stock, int cnt) {priority_queue<int, vector<int>, greater<int>> pq;for (auto e : stock) pq.push(e);vector<int> ret;while (cnt--){ret.push_back(pq.top());pq.pop();}return ret;}
};
快速选择算法。
我就说经常出错吧，这才几分钟…
else qsort(stock, right, r, cnt - a - b);
class Solution {
public:vector<int> inventoryManagement(vector<int>& stock, int cnt) {srand(time(nullptr));qsort(stock, 0, stock.size() - 1, cnt);return { stock.begin(), stock.begin() + cnt };}void qsort(vector<int>& stock, int l, int r, int cnt){if (l >= r) return;int key = stock[rand() % (r - l + 1) + l];int i = l, left = l - 1, right = r + 1;while (i < right){if (stock[i] < key) swap(stock[++left], stock[i++]);else if (stock[i] == key) i++;else swap(stock[--right], stock[i]);}int a = left - l + 1, b = right - left - 1;if (a >= cnt) qsort(stock, l, left, cnt);else if (a + b >= cnt) return;else qsort(stock, right, r, cnt - a - b);}
};
3、归并
排序数组
- 排序数组
class Solution {vector<int> tmp;
public:vector<int> sortArray(vector<int>& nums) {int sz = nums.size();tmp.resize(sz);MergeSort(nums, 0, sz - 1);return nums;}void MergeSort(vector<int>& arr, int l, int r){if (l >= r) return;int mid = (l + r) >> 1;MergeSort(arr, l, mid);MergeSort(arr, mid + 1, r);int b1 = l, b2 = mid + 1, i = l;while (b1 <= mid && b2 <= r){tmp[i++] = arr[b1] < arr[b2] ? arr[b1++] : arr[b2++];}while (b1 <= mid) tmp[i++] = arr[b1++];while (b2 <= r) tmp[i++] = arr[b2++];for (int i = l; i <= r; i++) arr[i] = tmp[i];}
};
LCR 170. 交易逆序对的总数
- LCR 170. 交易逆序对的总数
本题要求统计逆序对总数，笼统地说就是找前一部分的数有多少个数大于后一部分的数，那么对于前一部分和后一部分来说有序和乱序都是无所谓的。
排升序，找cur2之前有多少个数比我大。
class Solution {vector<int> tmp;
public:int reversePairs(vector<int>& record) {int sz = record.size();tmp.resize(sz);return MergeSort(record, 0, sz - 1);}int MergeSort(vector<int>& arr, int l, int r){if (l >= r) return 0;int mid = (l + r) >> 1;int ret = 0; // 逆序对个数ret += MergeSort(arr, l, mid);ret += MergeSort(arr, mid + 1, r);int b1 = l, b2 = mid + 1, i = l;while (b1 <= mid && b2 <= r){// 拍升序，找b2之前有多少个数比我大if (arr[b1] <= arr[b2]) tmp[i++] = arr[b1++];else{tmp[i++] = arr[b2++];ret += mid - b1 + 1;}}while (b1 <= mid) tmp[i++] = arr[b1++];while (b2 <= r) tmp[i++] = arr[b2++];for (int i = l; i <= r; i++) arr[i] = tmp[i];return ret;}
};
排降序，找cur1之后有多少个数比我小。
class Solution {vector<int> tmp;
public:int reversePairs(vector<int>& record) {int sz = record.size();tmp.resize(sz);return MergeSort(record, 0, sz - 1);}int MergeSort(vector<int>& arr, int l, int r){if (l >= r) return 0;int mid = (l + r) >> 1;int ret = 0; // 逆序对个数ret += MergeSort(arr, l, mid);ret += MergeSort(arr, mid + 1, r);int b1 = l, b2 = mid + 1, i = l;while (b1 <= mid && b2 <= r){// 排降序，找b1之后有多少个数比我小if (arr[b1] <= arr[b2]) tmp[i++] = arr[b2++];else{tmp[i++] = arr[b1++];ret += r - b2 + 1;}}while (b1 <= mid) tmp[i++] = arr[b1++];while (b2 <= r) tmp[i++] = arr[b2++];for (int i = l; i <= r; i++) arr[i] = tmp[i];return ret;}
};
计算右侧小于当前元素的个数
- 计算右侧小于当前元素的个数
class Solution {vector<int> ret;vector<int> index; // 绑定元素和其下标vector<int> tmp;vector<int> tmpindex;
public:vector<int> countSmaller(vector<int>& nums) {int n = nums.size();ret.resize(n);index.resize(n);tmp.resize(n);tmpindex.resize(n);for (int i = 0; i < n; i++) index[i] = i;MergeSort(nums, 0, n - 1);return ret;}void MergeSort(vector<int>& arr, int l, int r){if (l >= r) return;int mid = (l + r) >> 1;MergeSort(arr, l, mid);MergeSort(arr, mid + 1, r);int b1 = l, b2 = mid + 1, i = l;while (b1 <= mid && b2 <= r){if (arr[b1] <= arr[b2]) {tmp[i] = arr[b2];tmpindex[i++] = index[b2++]; // 下标和元素同步移动}else{ret[index[b1]] += r - b2 + 1;tmp[i] = arr[b1];tmpindex[i++] = index[b1++];}}while (b1 <= mid){tmp[i] = arr[b1];tmpindex[i++] = index[b1++];}while (b2 <= r){tmp[i] = arr[b2];tmpindex[i++] = index[b2++];}for (int i = l; i <= r; i++){arr[i] = tmp[i];index[i] = tmpindex[i];}}
};
翻转对
- 翻转对
class Solution {vector<int> tmp;
public:int reversePairs(vector<int>& nums) {int n = nums.size();tmp.resize(n);return MergeSort(nums, 0, n - 1);}int MergeSort(vector<int>& arr, int l, int r){if (l >= r) return 0;int mid = (l + r) >> 1;int ret = 0; // 统计翻转对个数ret += MergeSort(arr, l, mid);ret += MergeSort(arr, mid + 1, r);int b1 = l, b2 = mid + 1, i = l;// 和归并的逻辑不同，在归并之前统计while (b1 <= mid && b2 <= r){if (arr[b1] / 2.0 <= arr[b2]) b2++; // 做除法，防止溢出else{ret += r - b2 + 1;b1++;}}b1 = l, b2 = mid + 1;while (b1 <= mid && b2 <= r)tmp[i++] = arr[b1] <= arr[b2] ? arr[b2++] : arr[b1++];while (b1 <= mid) tmp[i++] = arr[b1++];while (b2 <= r) tmp[i++] = arr[b2++];for (int k = l; k <= r; k++) arr[k] = tmp[k];return ret;}
};
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