如果有一个字符的次数大于0了,说明现在滑动的位置已经缺字符了,又继续滑动右边的边界。 classSolution {public:stringminWindow(strings,stringt) { unordered_map<char,int>m;for(inti =0;i < t.size();i++) m[t[i]]++;stringres ="";intleft =0,count =0,length =INT_MAX;for(inti =0;i < s...
}privatebooleansearch(String s, TrieNode root,intindex,intnum) {if(root ==null)returnfalse;if(num > 0) {for(inti = 0; i < 26; i++) {if(search(s, root.children[i], index, num - 1))returntrue; }returnfalse; }if(index == s.length())returnroot.isWord;if(Character.isDigit(s...
Leetcode solution 243: Shortest Word Distance 2019-12-26 03:41 −Problem Statement Given a list of words and two words word1 and word2, return the shortest distance between these two word... 包子模拟面试 0 213 【leetcode】1289. Minimum Falling Path Sum II ...
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
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...
int[] res = new int[grid[0].length]; res[0] = grid[0][0]; for(int i=1;i<grid[0].length;i++) { res[i] = res[i-1]+grid[0][i]; } for(int i=1;i<grid.length;i++) { for(int j=0;j<grid[0].length;j++) ...
1publicclassSolution {2publicintminMutation(String start, String end, String[] bank) {3if(start==null|| end==null|| start.length()!=end.length())return-1;4intsteps = 0;5char[] mutations =newchar[]{'A', 'C', 'G', 'T'};6HashSet<String> validGene =newHashSet<String>();7fo...
the subarray [4,3] has the minimal length under the problem constraint. click to show more practice. 思路: 两个指针, start end, end向后走,直到 sum 大于 s. 然后start向后, 直到sum 小于s. 同时更新 min值。 AI检测代码解析 class Solution { ...
(":");int minutes=Integer.parseInt(time[0])*60+Integer.parseInt(time[1]);times[index++]=minutes;}Arrays.sort(times);int min=Integer.MAX_VALUE;int prev=times[times.length-1];for(int i=0;i<times.length;i++){int diff=(times[i]-prev);if(diff<0)diff+=1440;min=Math.min(min,diff...
int lenT = T.length(); 8 if (lenT <= 0) 9 return ""; 10 int n=0; 11 for (int i = 0; i < lenT; i++){ 12 int index = (int)(T.at(i)); 13 if (flag[index] == 0) 14 n++; 15 flag[index]++; 16 } 17 int alFlag[...