【LeetCode】496. Next Greater Element I

2019-11-10 20:16:31

字体：大中小

来源：转载

供稿：网友

【链接】：next-greater-element-I 【描述】：You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1’s elements in the corresponding places of nums2.

The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.

Example 1: Input: nums1 = [4,1,2], nums2 = [1,3,4,2]. Output: [-1,3,-1] Explanation: For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1. For number 1 in the first array, the next greater number for it in the second array is 3. For number 2 in the first array, there is no next greater number for it in the second array, so output -1. Example 2: Input: nums1 = [2,4], nums2 = [1,2,3,4]. Output: [3,-1] Explanation: For number 2 in the first array, the next greater number for it in the second array is 3. For number 4 in the first array, there is no next greater number for it in the second array, so output -1. Note: All elements in nums1 and nums2 are unique. The length of both nums1 and nums2 would not exceed 1000. 简单来说就是两个序列，求一个结果序列，第一个序列是第二个序列的子集且元素都unique，第一个序列的每个元素映射到第二个序列的相同的元素，如果第二个序列当前元素的右边出现比此元素大的元素则更新结果序列，否则更新为-1。【思路】第二种解法容易想到。第一种解法其实就是多一个search函数,每次查询findNums的元素，每次更新即可。【代码】：

/***********************【LeetCode】496. Next Greater Element IAuthor：herongweiTime:2017/2/8 11:50language：C++http://blog.csdn.net/u013050857***********************/#PRagma comment(linker,"/STACK:102400000,102400000")#include <bits/stdc++.h>#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;typedef long long LL;const int maxn = 1e5+10;const int maxm = 55;const LL MOD = 999999997;int dir4[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};int dir8[8][2]= {{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};inline LL read(){ int c=0,f=1; char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){c=c*10+ch-'0';ch=getchar();} return c*f;}class Solution1{public: vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) { int findNumsSize=findNums.size(); int numsSize=nums.size(); vector<int> ret(findNumsSize); for(int i=0; i<findNumsSize; ++i){ ret[i]=search(nums,findNums[i]); } return ret; } int search(vector<int>& nums,int key) { int i; for(i=0; i<nums.size(); ++i){ if(nums[i]==key) break; } if(i==nums.size()||i==nums.size()-1) return -1; for(;i<nums.size(); ++i) { if(nums[i]>key) return nums[i]; } return -1; }};class Solution2{ public: vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) { int findNumsSize=findNums.size(); int numsSize=nums.size(); unordered_map<int ,int >ans; vector<int> ret; for(int i=0; i<numsSize; ++i){ ans[nums[i]]=i+1; } for(int i=0; i<findNumSize; ++i){ int j=ans[findNums[i]]; while(j<numsSize&&findNums[i]>=nums[j]) ++j;//findNums[i]>=nums[j]右边第一个大的数！ if(j<numsSize) ret.push_back(nums[j]); else ret.push_back(-1); } return ret; }};

上一篇：hardfault排故记录01

下一篇：ZJOI 2008 瞭望塔三分法