【鏈接】：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. 簡單來說就是兩個序列，求一個結(jié)果序列，第一個序列是第二個序列的子集且元素都unique，第一個序列的每個元素映射到第二個序列的相同的元素，如果第二個序列當前元素的右邊出現(xiàn)比此元素大的元素則更新結(jié)果序列，否則更新為-1。 【思路】第二種解法容易想到。第一種解法其實就是多一個search函數(shù),每次查詢findNums的元素，每次更新即可。 【代碼】：
/***********************【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]右邊第一個大的數(shù)！ if(j<numsSize) ret.push_back(nums[j]); else ret.push_back(-1); } return ret; }};新聞熱點
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