下面是我写的,需要的时间复杂度为O(length(S)),即O(n),空间复杂度为O(length(T)):View Codeleetcode 讨论里面有另外一个实现,也是O(n),实现类似,不同的是循环内是通过判断窗口中已经有T中字符的个数来进行 add or delete,我写的是通过目标操作字符来进行,逻辑上没有ta写的方便。https://oj.leetcode....
Given an array of positive integers nums and a positive integer target, return the minimal length of a contiguous subarray [numsl, numsl+1, ..., numsr-1, numsr] of which the sum is greater than or equal to target. If there is no such subarray, return 0 instead....
github:https://github.com/ChopinXBP/LeetCode-Babel 二维数组路径题很容易可以想到用回溯的方法,不过这题回溯会报超时。从题目限定的向右和向下的特点出发,可以从起点出发采用动态规划的方法求解起点至终点的矩阵每一点的最小路径和。 dp[i][j]代表起点至ij点的路径和,状态转移分为四种情况: 1.起点的最小路径...
输入:s = "ADOBECODEBANC", t = "ABC" 输出:"BANC" 解释:最小覆盖子串 "BANC" 包含来自字符串 t 的 'A'、'B' 和 'C'。 示例2: 输入:s = "a", t = "a" 输出:"a" 解释:整个字符串 s 是最小覆盖子串。 示例3: 输入: s = "a", t = "aa" 输出: "" 解释: t 中两个字符 '...
在 "leet" 中删除 "e" 将 101[e] 加入总和。 结束时,两个字符串都等于 "let",结果即为 100+101+101+101 = 403 。 如果改为将两个字符串转换为 "lee" 或 "eet",我们会得到 433 或 417 的结果,比答案更大。 提示: 0 <= s1.length, s2.length <= 1000...
Output: 2 Explanation: The only solution is to delete the first two characters. 1. 2. 3. Constraints: 1 <= s.length <= 105 s[i] is ‘a’ or 'b’. 分析 题目的意思是:给定一个字符串s,删除其中的字符,使得字符串平衡,即前半部分为全a,后半部分为全b。
MAX_VALUE; int[] sums = new int[length + 1]; for (int i = 1; i <= length; i++) { sums[i] = sums[i - 1] + nums[i - 1]; } for (int i = 0; i <= length; i++) { int target = s + sums[i]; int index = Arrays.binarySearch(sums, target); if (index < 0)...
given a fixed length K of a sliding window. so we are given an array as well. and this sliding window slides all the way, and get the maximum each time it moves. return the results in array. my idea: wow, it looks like a deque and priority queue problem! and I remember the imple...
If instead we turned both strings into “lee” or “eet”, we would get answers of 433 or 417, which are higher. Note: 0 < s1.length, s2.length <= 1000. All elements of each string will have an ASCII value in [97, 122]. 思路: 动态规划,和最小编辑距离相似,先用递归的方式解一遍...
publicStringminWindow(Strings,Stringt){Map<Character,Integer>map=newHashMap<>();//遍历字符串 t,初始化每个字母的次数for(inti=0;i<t.length();i++){charchar_i=t.charAt(i);map.put(char_i,map.getOrDefault(char_i,0)+1);}intleft=0;//左指针intright=0;//右指针intans_left=0;//保存...