LeetCode[421] Maximum XOR of Two Numbers in an Array Given a non-emptyarrayof numbers, a0, a1, a2, … , an-1, where 0 ≤ ai < 231. Find the maximum result of ai XOR aj, where 0 ≤ i, j < n. Could you do this in O(n) runtime? Example: Input: [3, 10, 5, 25, 2...
链接:https://leetcode.com/problems... 可以用trie tree来做。把所有num放进tree,之后对每一个num从最高位找是否存在和当前bit不同的path。把build tree单独写一个函数,结果TLE了,所以放一起了。。 public class Solution { public int findMaximumXOR(int[] nums) { /* trie tree: root is the largest...
// 非常非常棒 // 参考了 https://discuss.leetcode.com/topic/63213/java-o-n-solution-using-bit-manipulation-and-hashmap // 特别的,利用了异或的强大运算特性,见22行,来加速运算 public class Solution { public int findMaximumXOR(int[] nums) { int max = 0; int flag = 0; // from left to...
421. Maximum XOR of Two Numbers in an Array Given a non-empty array of numbers, a0, a1, a2, … , an-1,where0≤ ai <231. Find the maximum result of ai XOR aj,where0≤ i, j <n. Could youdothisinO(n) runtime?Example: Input: [3,10,5,25,2,8] Output:28Explanation: The m...
curNode= curNode.children[curBit ^ 1]; }else{ curNode=curNode.children[curBit]; } } max=Math.max(curSum, max); }returnmax; } } https://leetcode.com/problems/maximum-xor-of-two-numbers-in-an-array/discuss/91059/Java-O(n)-solution-using-Trie...
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// Utility function to find the maximum of two numbers intmax(intx,inty){ return(x>y)?x:y; } // Function to find the maximum subarray sum using divide-and-conquer intmaximum_sum(intnums[],intlow,inthigh) { // If the array contains 0 or 1 element ...
Given a non-empty array of numbers, a0, a1, a2, … , an-1, where 0 ≤ ai < 231. Find the maximum result of ai XOR aj, where0 ≤ i, j <n. Could youdothisin O(n) runtime?Example: Input: [3, 10, 5, 25, 2, 8] ...
leetcode 628. Maximum Product of Three Numbers(三个数的最大乘积)--Java题解,程序员大本营,技术文章内容聚合第一站。
Both numbers with value 2 are both considered as second maximum. */ publicclassThirdMaximumNumber { /** * Main method *@paramargs *@throwsException */ publicstaticvoidmain(String[]args)throwsException { int[] a={1,2}; System.out.println(newThirdMaximumNumber().thirdMax(a)); ...