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brute force O(n2)从头尾两头扫面 O(n) 1 public class Solution { 2 public int maxArea(int[] height) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 int start = 0; 6 int end = height.length... 阅读全文
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注意负数-2147483648的绝对值还是自己！！！(overflow) 1 public class Solution { 2 public boolean isPalindrome(int x) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 if( x 0)11 {12 mask... 阅读全文
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1.trim2.符号3.overflow 1 public class Solution { 2 public int atoi(String str) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 if(str == null) 6 return 0; 7 if(str.lengt... 阅读全文
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use PriorityQueue( Priority Heap)the time complexity will be O(k * lgn) 1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * }10 * }11 */12 public cla... 阅读全文
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用stack，并记录入栈的括号的位置 1 public class Solution { 2 class Node{ 3 char c; 4 int nub; 5 public Node(char c, int nub){ 6 this.c = c; 7 this.nub = nub; 8 } 9 }10 11 public int longestValidParentheses(String s) {12 // IMPORTANT: ... 阅读全文
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两次二分法查找 1 public class Solution { 2 public boolean searchMatrix(int[][] matrix, int target) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 int line = searchLine(matrix,0,matrix.length,ta... 阅读全文
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直接乘的话会超时O(n)。利用二分法降为O(lgn)。 1 public class Solution { 2 public double pow(double x, int n) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 if(n == 0) 6 return 1; 7 if(... 阅读全文
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If result is overflow, then return -1. 1 public class Solution { 2 public int reverse(int x) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 boolean positive = true; 6 if(x 0)11 ... 阅读全文
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寻找W型字符的规律：除第一和最后一行外都是对称的。找出循环的长度相应的做出位移即可。比较烦 1 public class Solution { 2 public String convert(String s, int nRows) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 if(nRows == 1) 6 ... 阅读全文
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从左往右扫描，当遇到重复字母时，以上一个重复字母的index +1，作为新的搜索起始位置。可以减少时间， 但时间复杂度没有变O(n2) 1 public class Solution { 2 public int lengthOfLongestSubstring(String s) { 3 // IMPORTANT: Please reset any member data you declared, as 4 // the same Solution instance will be reused for each test case. 5 ... 阅读全文
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