summer_zhou

摘要： void connect(TreeLinkNode *root) { // Note: The Solution object is instantiated only once and is reused by each test case. if(!root) return; TreeLinkNode* first,*cur; first = cur = root; while(cur) { if(cur->left) { ... 阅读全文

摘要： vector getRow(int rowIndex) { // Note: The Solution object is instantiated only once and is reused by each test case. if(rowIndex(); vector res(rowIndex+1); res[0] = 1; for(int i=1;i0;j--) res[j] += res[j-1]; res[0] = 1; ... 阅读全文

摘要： //注意容器的空间，如果没有预设好，[]访问会出错的。。。。---这个很容易出问题。 vector > generate(int numRows) { // Note: The Solution object is instantiated only once and is reused by each test case. vector> res; if(numRows()); res[0].push_back(1); for(int i=1;i<numRows;i++) { ... 阅读全文

摘要： ListNode *deleteDuplicates(ListNode *head) { // Note: The Solution object is instantiated only once and is reused by each test case. if(head==NULL) return NULL; ListNode* prev = head,*cur = head->next; while(cur) { ListNode* next =... 阅读全文

摘要： DP。O(n)的空间复杂度。时间复杂度O(n^2)sum[i][j] = min(sum[i-1][j-1],sum[i-1][j])+triangle[i][j]; //所该式可以看出，可以用O(n)的额外空间。求解的时候，记得从大往小求。 int minimumTotal(vector > &triangle) { // Note: The Solution object is instantiated only once and is reused by each test case. if(triangle.empty()) re... 阅读全文

摘要： 暴力法即可。不需要用kmp，太复杂了。 char *strStr(char *haystack, char *needle) { // Note: The Solution object is instantiated only once and is reused by each test case. if(!needle) return haystack; if(!haystack) return NULL; int len1 = strlen(haystack); ... 阅读全文

摘要： 考虑负数做负号运算符之后的溢出问题。-INT_MIN int已经无法表示，溢出了，所以用long long来存int divide(int dividend, int divisor) { // Note: The Solution object is instantiated only once and is reused by each test case. if(divisor==0) return 0; bool bNega = false; long long divid,divis; if((dividend>0&&divisor0)) bNega = tru 阅读全文

摘要： ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { // Start typing your C/C++ solution below // DO NOT write int main() function int carry = 0; ListNode* nhead=NULL,*prev=NULL; ListNode* it1=l1,*it2=l2; while(it1||it2||carry) { int n... 阅读全文

摘要： 注意边界条件即可：num1表示0或者num2表示0； string multiply(string num1, string num2) { // Note: The Solution object is instantiated only once and is reused by each test case. if(num1.empty()||num2.empty()) return ""; if(num1[0]=='0'||num2[0]=='0') //ATT here return "0";... 阅读全文

摘要： string addBinary(string a, string b) { // Note: The Solution object is instantiated only once and is reused by each test case. int m = a.size(); int n = b.size(); string res(max(m,n)+1,'0'); int i = m-1,j=n-1; int carry = 0; int k = res.si... 阅读全文