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....
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....
costs.length == 3 1 <= costs[i] <= 1000 Choose a type Comment 💡 Discussion Rules 1. Please don't postany solutionsin this discussion. 2. The problem discussion is for asking questions about the problem or for sharing tips - anything except for solutions. ...
1247. Minimum Swaps to Make Strings Equal** https://leetcode.com/problems/minimum-swaps-to-make-strings-equal/ 题目描述 You are given two strings s1 and s2 of equal length consisting of letters "x&qu...Minimum Moves to Equal leetcode Minimum Moves to Equal问题 邮局问题,找到中位数即可 ...
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。
There are two possible trees. The first has non-leaf node sum 36, and the second has non-leaf node sum 32. 24 24 / \ / \ 12 4 6 8 / \ / \ 6 2 2 4 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Constraints: 2 <= arr.length <= 40 ...
classSolution{public intminimumCost(int[]cost){int n=cost.length,res=0;Arrays.sort(cost);for(int i=0;i<n;i++){if(i%3!=n%3){res+=cost[i];}}returnres;}} Runtime:2 ms, faster than100.00%of Java online submissions for Minimum Cost of Buying Candies With Discount. ...
注意最后需要判断length是否为math.MaxInt32 若是则表示没有匹配的字符返回空串 若不是则直接返回对应子串 代码 func minWindow(s string, t string) string { need,windows:=make(map[byte]int,0),make(map[byte]int,0) for _,v:=range t{ need[byte(v)]++ } l,r,length,start,valid:=0,0,math...
贪心解。尽可能的选择最前面的两个不相同的元素，得到的就是最长的美丽数组。然后用总长度减去该长度即为答案。还有
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...