532. K-diff Pairs in an Array\531. Lonely Pixel I

532. K-diff Pairs in an Array

题目描述

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.

Example 1:Input: [3, 1, 4, 1, 5], k = 2Output: 2Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).Although we have two 1s in the input, we should only return the number of unique pairs.Example 2:Input:[1, 2, 3, 4, 5], k = 1Output: 4Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).Example 3:Input: [1, 3, 1, 5, 4], k = 0Output: 1Explanation: There is one 0-diff pair in the array, (1, 1).Note:The pairs (i, j) and (j, i) count as the same pair.The length of the array won't exceed 10,000.All the integers in the given input belong to the range: [-1e7, 1e7].

就是在一个序列中，计算序列中各个元素的差的绝对值，然后找到与目标差的绝对值相等的个数。其中要注意的是相差的值为的一对数不能重复。

代码实现

法一：
直接上暴力破解。

class Solution {public:    int findPairs(vector<int>& nums, int k) {        int len = nums.size(), res = 0;        if(k < 0) return res;        sort(nums.begin(), nums.end());        for(int i = 0; i < len; i++) {            if(i > 0 && nums[i] == nums[i-1]) continue;            int d1 = nums[i] - k, d2 = nums[i] + k;            for(int j = i+1; j < len; j++) {                if(j > i+1 && nums[j] == nums[j-1]) continue;                if(nums[j] == d1 || nums[j] == d2) res++;            }        }        return res;    }};

法二：使用set和map一起协作。

使用map记录出现过的数字，使用set记录一对符合要求的数的更小数。

class Solution {public:    int findPairs(vector<int>& nums, int k) {        int len = nums.size(), res = 0;        if(k < 0) return res;        set<int> st;        map<int, int> mp;        for(int i = 0; i < len; i++) {            int d1 = nums[i] + k, d2 = nums[i] - k;            if(mp.count(nums[i]+k)) {  st.insert(nums[i]); }            if(mp.count(d2)) { st.insert(d2); }            mp[nums[i]]++;        }        return st.size();    }};

531. Lonely Pixel I

题目描述

Given a picture consisting of black and white pixels, find the number of black lonely pixels.

The picture is represented by a 2D char array consisting of ‘B’ and ‘W’, which means black and white pixels respectively.

A black lonely pixel is character ‘B’ that located at a specific position where the same row and same column don’t have any other black pixels.

Example:

Input: [['W', 'W', 'B'], ['W', 'B', 'W'], ['B', 'W', 'W']]

Output: 3
Explanation: All the three ‘B’s are black lonely pixels.

Note:
The range of width and height of the input 2D array is [1,500].

这道题目就是说如果一个字符B在横纵都是只有一个，那么我们认为结果就是一个lonely的数，然后统计有多少个这样的结果。

代码实现

使用哈希统计横纵的字符B的个数，然后取更小的数值为1的数。

class Solution {public:    int findLonelyPixel(vector<vector<char>>& pic) {        int m = pic.size(), n = m?pic[0].size():0, res = 0;        if(!m) return 0;        cout << m<< "  "<<n<< endl;        vector<int> hash(m+n, 0);        for(int i = 0; i < m; i++) {            for(int j = 0; j < n; j++) {                if(pic[i][j] == 'B') {                    hash[i]++;                    hash[m+j]++;                }            }        }        int cnt1 = 0, cnt2 = 0;        for(int i = 0; i < m; i++)            if(hash[i] == 1) cnt1++;        for(int i = 0; i < n; i++)            if(hash[i+m] == 1) cnt2++;        return min(cnt1, cnt2);    }};