228. Summary Ranges

最新推荐文章于 2024-10-31 00:15:00 发布
原创 最新推荐文章于 2024-10-31 00:15:00 发布 · 189 阅读 · 0 ·
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class Solution {
public:
    vector<string> summaryRanges(vector<int>& nums) {
        int n=nums.size(),i;
        vector<string> ranges;
        if(n==0)
            return ranges;
            
        string tmp="";
        int cnt=0;
        for(i=0;i<n;)
        {
            if(tmp=="")
            {
                tmp=to_string(nums[i]);
                cnt=1;
            }
            while(i<n-1&&nums[i]+1==nums[i+1])
            {
                cnt++;
                i++;
            }
            if(cnt==1)
                ranges.push_back(tmp);
            else
            {
                tmp+="->";
                tmp+=to_string(nums[i]);
                ranges.push_back(tmp);
            }
            tmp="";
            i++;
        }
        return ranges;
    }
};

LeetCode --- 228. Summary Ranges 解题报告
杨鑫newlife的专栏
07-26 392
You are given asorted uniqueinteger arraynums. Returnthesmallest sortedlist of ranges thatcover all the numbers in the array exactly. That is, each element ofnumsis covered by exactly one of the ranges, and there is no integerxsuch thatxis in...
别把数组当“流水账”：用 Summary Ranges 教你学会“结构化思考”
Echo_Wish
12-12 15
这篇文章通过讲解“汇总区间”（Summary Ranges）算法，揭示了结构化思维的重要性。作者指出，这个看似简单的算法实际上体现了将线性数据转化为段落式信息的能力，就像写文章需要分段一样。文章详细解析了算法原理，包括如何维护区间起点终点、何时中断分段等关键点，并提供了Python实现代码。更重要的是，作者通过日志分析、运维监控等实际场景，展示了该算法的广泛应用价值。最后升华主题，指出这种“分段”思维不仅适用于编程，也是处理复杂问题、规划人生的通用逻辑——学会识别连续性、突变点，并适时收口重组信息。全文以生
228.Summary Ranges
dashi7657381的博客
04-11 183
Input:[0,1,2,4,5,7]Output:["0->2","4->5","7"]Explanation:0,1,2 form a continuous range;4,5 form a continuous range. class Solution { public List<String> summaryRanges(int[] ...
LeetCode - 228. Summary Ranges
virgilshi的博客
07-23 194
题目 LeetCode - 228. Summary Ranges 题目链接 https://leetcode.com/problems/summary-ranges/ 参考博客 [1] https://blog.csdn.net/hlsdbd1990/article/details/46311137 解题思路 用一个容器将属于一个范围的数据存起来，然后统一转换。 额外发现： STL容器中只有ve...
LeetCode C++ 228. Summary Ranges【Array】简单
memcpy0的博客
01-10 422
You are given a sorted unique integer array nums. Return the smallest sorted list of ranges that cover all the numbers in the array exactly. That is, each element of nums is covered by exactly one of the ranges, and there is no integer x such that x is in
python leetcode 228. Summary Ranges
Neekity的博客
12-18 499
class Solution: def summaryRanges(self, nums): """ :type nums: List[int] :rtype: List[str] """ if not nums: return [] n = len(nums) ...
228. Summary Ranges 题解
yijingzhi的博客
06-08 305
228. Summary Ranges 题解 题目描述： Given a sorted integer array without duplicates, return the summary of its ranges. For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"].
Leetcode 228. Summary Ranges
小白菜又菜
04-23 158
【代码】Leetcode 228. Summary Ranges。
闯关leetcode——228. Summary Ranges
方亮的专栏
10-31 1005
这题就是要将一个从小到大排序的数组，使用一组区间表示。孤立的区间只记录孤立的值，长度超过1的区间使用“起始值->结束值”的方式表示。解题思路就是记录起始值后，向后寻找连续的最终值。根据区间中元素个数决定表达方式。
python-leetcode题解之228-Summary-Ranges.py
09-12
汇总区间”（Summary Ranges）的Python 解决方案。 “228. 汇总区间”是一道中等难度的题目，题目要求编写一个函数，输入一个无序的整数数组，输出该数组中所有区间合并后的结果。具体要求是，如果一组数字连续，...
由于数据清洗需要，我需要一个python脚本工具批量处理。我编写了代码如下： import os import pandas as pd from openpyxl import load_workbook from openpyxl.utils import get_column_letter from openpyxl.worksheet.merge import MergeCell def clean_excel_files(input_dir, output_dir, summary_keywords, target_content, delete_type='row'): os.makedirs(output_dir, exist_ok=True) for filename in os.listdir(input_dir): if filename.endswith(('.xlsx', '.xls')): input_path = os.path.join(input_dir, filename) output_path = os.path.join(output_dir, filename) wb = load_workbook(input_path) for sheet_name in wb.sheetnames: ws = wb[sheet_name] # 1. 删除汇总行 rows_to_delete = [] for row in ws.iter_rows(): for cell in row: if any(keyword in str(cell.value) for keyword in summary_keywords): rows_to_delete.append(cell.row) break for row_idx in sorted(set(rows_to_delete), reverse=True): ws.delete_rows(row_idx) # 2. 处理目标内容（改进合并单元格处理） if delete_type == 'row': # 存储待删除的行索引（使用集合避免重复） rows_to_delete_target = set() # 处理合并单元格区域 if ws.merged_cells.ranges: merged_ranges_to_remove = [] # 存储需要移除的合并区域 for merged_range in list(ws.merged_cells.ranges): top_left_cell = ws.cell(merged_range.min_row, merged_range.min_col) if top_left_cell.value == target_content: # 记录合并区域覆盖的所有行 for r in range(merged_range.min_row, merged_range.max_row + 1): rows_to_delete_target.add(r) # 标记该合并区域需要移除 merged_ranges_to_remove.append(merged_range) # 移除包含目标内容的合并区域 for merged_range in merged_ranges_to_remove: ws.unmerge_cells(str(merged_range)) # 处理非合并单元格 for row in ws.iter_rows(): for cell in row: # 只处理未在合并区域中处理过的单元格 if cell.value == target_content and cell.row not in rows_to_delete_target: rows_to_delete_target.add(cell.row) # 执行行删除 for row_idx in sorted(rows_to_delete_target, reverse=True): ws.delete_rows(row_idx) else: # 列删除逻辑保持不变 cells_to_delete = [] for row in ws.iter_rows(): for cell in row: if cell.value == target_content: cells_to_delete.append(cell.column) if cells_to_delete: for col_idx in sorted(set(cells_to_delete), reverse=True): ws.delete_cols(col_idx) wb.save(output_path) print(f"文件已处理: {filename}") # 使用示例 if __name__ == "__main__": input_directory = "C:\\Users\\结算风控部\\Desktop\\FuturesRebateSystem\\data\\input" output_directory = "C:\\Users\\结算风控部\\Desktop\\FuturesRebateSystem\\data\\cleaned" summary_keywords = ["总计", "合计", "汇总", "小计", "共计", "制表人","深圳市","日期"] target_content = "投资者资金对账" delete_type = "row" clean_excel_files(input_directory, output_directory, summary_keywords, target_content, delete_type) 现在要求先解除合并单元格区域再执行删除操作。请在我的代码基础上进行改进
最新发布
12-17
merged_ranges = list(ws.merged_cells.ranges) # 获取所有合并区域 for merged_range in merged_ranges: min_row, min_col, max_row, max_col = range_boundaries(str(merged_range)) top_left_value = ws.cell...
import os import pandas as pd from openpyxl import load_workbook from openpyxl.utils import get_column_letter from openpyxl.worksheet.merge import MergeCell def clean_excel_files(input_dir, output_dir, summary_keywords, target_content, delete_type='row'): os.makedirs(output_dir, exist_ok=True) for filename in os.listdir(input_dir): if filename.endswith(('.xlsx', '.xls')): input_path = os.path.join(input_dir, filename) output_path = os.path.join(output_dir, filename) wb = load_workbook(input_path) for sheet_name in wb.sheetnames: ws = wb[sheet_name] # 1. 删除汇总行 rows_to_delete = [] for row in ws.iter_rows(): for cell in row: if any(keyword in str(cell.value) for keyword in summary_keywords): rows_to_delete.append(cell.row) break for row_idx in sorted(set(rows_to_delete), reverse=True): ws.delete_rows(row_idx) # 2. 处理目标内容（改进合并单元格处理） if delete_type == 'row': # 存储待删除的行索引（使用集合避免重复） rows_to_delete_target = set() # 处理合并单元格区域 if ws.merged_cells.ranges: merged_ranges_to_remove = [] # 存储需要移除的合并区域 for merged_range in list(ws.merged_cells.ranges): top_left_cell = ws.cell(merged_range.min_row, merged_range.min_col) if top_left_cell.value == target_content: # 记录合并区域覆盖的所有行 for r in range(merged_range.min_row, merged_range.max_row + 1): rows_to_delete_target.add(r) # 标记该合并区域需要移除 merged_ranges_to_remove.append(merged_range) # 移除包含目标内容的合并区域 for merged_range in merged_ranges_to_remove: ws.unmerge_cells(str(merged_range)) # 处理非合并单元格 for row in ws.iter_rows(): for cell in row: # 只处理未在合并区域中处理过的单元格 if cell.value == target_content and cell.row not in rows_to_delete_target: rows_to_delete_target.add(cell.row) # 执行行删除 for row_idx in sorted(rows_to_delete_target, reverse=True): ws.delete_rows(row_idx) else: # 列删除逻辑保持不变 cells_to_delete = [] for row in ws.iter_rows(): for cell in row: if cell.value == target_content: cells_to_delete.append(cell.column) if cells_to_delete: for col_idx in sorted(set(cells_to_delete), reverse=True): ws.delete_cols(col_idx) wb.save(output_path) print(f"文件已处理: {filename}") # 使用示例 if __name__ == "__main__": input_directory = "C:\\Users\\结算风控部\\Desktop\\FuturesRebateSystem\\data\\input" output_directory = "C:\\Users\\结算风控部\\Desktop\\FuturesRebateSystem\\data\\cleaned" summary_keywords = ["总计", "合计", "汇总", "小计", "共计", "制表人","深圳市","日期"] target_content = "投资者资金对账" delete_type = "row" clean_excel_files(input_directory, output_directory, summary_keywords, target_content, delete_type) 现在target_content所在单元格为合并单元格，执行删除并未解除合并单元格（导致后续数据被挤掉只显示第一个单元格的值），请改进代码
12-17
merged_ranges = list(ws.merged_cells.ranges) # 1. 处理目标区域内的合并单元格 affected_merges = [] for merge in merged_ranges: min_row, min_col, max_row, max_col = merge.min_row, merge.min_col, ...
73. Set Matrix Zeroes
varcom的博客
09-02 1115
方法一 将第一行和第一列作为标记。matrix[i][j]==0,将matrix[i][0]和matrix[0][j]设为0，所以matrix[0][0]要单独考虑，也要最后遍历 class Solution { public: void setZeroes(vector>& matrix) { int n=matrix.size(); if(n==0)
48. Rotate Image
varcom的博客
09-01 872
转置后的矩阵最外围的元素不变，可以根据这个性质递归程序 class Solution { public: void rotate(vector>& matrix,int x,int y,int m,int n) { if(!(x<m&&y<n)) return; swap(matrix[x][y],matrix[m][n])
381. Insert Delete GetRandom O(1) - Duplicates allowed
varcom的博客
09-07 715
使用unordered_map和unordered_set存取 class RandomizedCollection { public: /** Initialize your data structure here. */ RandomizedCollection() { mp.clear(); } /** Inserts a valu
16. 3Sum Closest
varcom的博客
08-29 442
枚举＋二分搜索 class Solution { public: int threeSumClosest(vector& nums, int target) { int n=nums.size(); int i,j,k; int ans=nums[0]+nums[1]+nums[2]; sort(nums.begin(),n
18. 4Sum
varcom的博客
08-29 389
枚举第一个数和最后一个数，中间两个数二分查找 class Solution { public: vector> fourSum(vector& nums, int target) { int n=nums.size(); int i,j,k,t; sort(nums.begin(),nums.end());
26. Remove Duplicates from Sorted Array
varcom的博客
08-29 333
class Solution { public: int removeDuplicates(vector& nums) { int n=nums.size(); int i,k=0,length=n; for(i=1;i<n;i++) { if(nums[i]==nums[i-1-k])
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