C void dfs(int pos, int orVal, const int* nums, int numsSize, int* maxOr, int* cnt) { if (pos == numsSize) { if (orVal > *maxOr) { *maxOr = orVal; *cnt = 1; } else if (orVal == *maxOr) { (*cnt)++; } return; } dfs(pos + 1, orVal | nums[pos], nums, ...
max(maxCnt, c); } return wall.length - maxCnt; }; Golang func leastBricks(wall [][]int) int { cnt := map[int]int{} for _, widths := range wall { sum := 0 for _, width := range widths[:len(widths)-1] { sum += width cnt[sum]++ } } maxCnt := 0 for _, c :...
上一题:LeetCode第10题: isMatch(C语言) 思路: 1、基本解法 思路:建立两层循环,分别计算数组两个元素的乘积,与记录的最大值比较 intmaxArea(int*height,intheightSize){intmax_area=0;for(inti=0;i<heightSize;i++){for(intj=heightSize-1;j>=0&&j>i;j--){intmin_height=0;if(height[i]-height...
此时我们遍历到了坡顶， 然后就要下坡，此时记录当前坡顶分发糖果数num，之后遍历至坡底，同理坡底赋值1，然后反向回来找刚刚的 坡顶，每次candy[i] = candy[i+1] + 1，到坡顶后有candy[i] = max(candy[i],num)。结束后candy即为 答案，最后相加即可。 然后写的时候发现很难写 又...
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其实也很简单,组合数学。将数组切分成 D 个区间,不同的切分方式为不同的方案。最优方案是 min(max(切分区间))min(max(切分区间)),复杂度为 C(n−1,D)C(n−1,D), 阶乘级别。而我们根据题目中隐含的单调性将复杂度极大地降低,下面我们来看看同类型的其他题目,首先给出个定义。在值域上二分的这类...
res=max(res, cnt); } }for(intoffset =0; offset < n; ++offset) {for(inti =0, j = offset; i < m && j <n;) {intcnt =0;while(i < m && j < n && A[i++] == B[j++]) ++cnt; res=max(res, cnt); } }returnres; ...
functionleastInterval(tasks,n){letarr=Array(26).fill(0);for(letcoftasks){//统计各个字母出现的次数arr[c.charCodeAt()-"A".charCodeAt()]++;}letmax=0;for(leti=0;i<26;i++){//找到最大次数max=Math.max(max,arr[i]);}letret=(max-1)*(n+1);//计算前n-1行n的间隔的时间大小for(leti...
//priority_queue <int,vector<int>,greater<int> > que(k, INT_MAX); //降序队列,大顶堆 //priority_queue <int,vector<int>,less<int> >que(k, INT_MIN); //greater和less是std实现的两个仿函数(就是使一个类的使用看上去像一个函数。其实现就是类中实现一个operator(),这个类就有了类似函数的...
1593.Split-a-String-Into-the-Max-Number-of-Unique-Substrings (M) 1681.Minimum-Incompatibility (H) 1723.Find-Minimum-Time-to-Finish-All-Jobs (H-) 2305.Fair-Distribution-of-Cookies (H-) 2597.The-Number-of-Beautiful-Subsets (M+) 2842.Count-K-Subsequences-of-a-String-With-Maximum-Beauty ...