数组类，简单级别完结。。。。
不容易啊，基本都是靠百度答案。。。。
希望做过之后后面可以自己复习，自己学会这个解法
package y2019.Algorithm.array;/**
* @ProjectName: cutter-point
* @Package: y2019.Algorithm.array
* @ClassName: CanPlaceFlowers
* @Author: xiaof
* @Description: TODO 605. Can Place Flowers
* Suppose you have a long flowerbed in which some of the plots are planted and some are not.
* However, flowers cannot be planted in adjacent plots - they would compete for water and both would die.
* Given a flowerbed (represented as an array containing 0 and 1, where 0 means empty and 1 means not empty),
* and a number n, return if n new flowers can be planted in it without violating the no-adjacent-flowers rule.
*
* 假设你有一个很长的花坛，一部分地块种植了花，另一部分却没有。可是，花卉不能种植在相邻的地块上，它们会争夺水源，两者都会死去。
* 给定一个花坛（表示为一个数组包含0和1，其中0表示没种植花，1表示种植了花），和一个数 n 。能否在不打破种植规则的情况下种入 n 朵花？能则返回True，不能则返回False。
* 来源：力扣（LeetCode）
* 链接：https://leetcode-cn.com/problems/can-place-flowers
* 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
*
* Input: flowerbed = [1,0,0,0,1], n = 1
* Output: True
* @Date: 2019/7/12 8:55
* @Version: 1.0
*/
public class CanPlaceFlowers { public boolean solution(int[] flowerbed, int n) {
//判断每个花之间要有空格
int count = 0;
for(int i = 0; i < flowerbed.length && count < n; ++i) {
//判断是否是空的
if(flowerbed[i] == 0) {
//判断左右是否间隔1一个空位
int pre = i == 0 ? 0 : flowerbed[i - 1];
int next = i < flowerbed.length - 1 ? flowerbed[i + 1] : 0;
if(pre == 0 && next == 0) {
++count;
flowerbed[i] = 1;
}
}
} if(count == n) {
return true;
} else {
return false;
}
} public static void main(String args[]) {
String as[] = {"bella","label","roller"};
int[] A = {1,0,0,0,1};
int k = 1;
CanPlaceFlowers fuc = new CanPlaceFlowers();
System.out.println(fuc.solution(A, k));
}
}
package y2019.Algorithm.array;import java.util.HashMap;
import java.util.Map;/**
* @ProjectName: cutter-point
* @Package: y2019.Algorithm.array
* @ClassName: FindPairs
* @Author: xiaof
* @Description: 532. K-diff Pairs in an Array
* 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.
*
* Input: [3, 1, 4, 1, 5], k = 2
* Output: 2
* Explanation: 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.
*
* 给定一个整数数组和一个整数 k, 你需要在数组里找到不同的 k-diff 数对。这里将 k-diff 数对定义为一个整数对 (i, j),
* 其中 i 和 j 都是数组中的数字，且两数之差的绝对值是 k.
*
* 来源：力扣（LeetCode）
* 链接：https://leetcode-cn.com/problems/k-diff-pairs-in-an-array
* 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
*
* @Date: 2019/7/12 9:33
* @Version: 1.0
*/
public class FindPairs { public int solution(int[] nums, int k) { if(k < 0) {
return 0;
} //这题用类似hash的方式
Map<Integer, Integer> numMap = new HashMap();
for(int t : nums) {
if(!numMap.containsKey(t)) {
numMap.put(t, 0);
}
//出现次数
numMap.put(t, numMap.get(t) + 1);
}
//遍历查询缺失的一般
int count = 0;
//根据map中的数据进行判断，并且进行了去重
for(Map.Entry entry : numMap.entrySet()) {
//判断map中是否包含差值的数据，如果包含，还要避免是同同一对数据,剩余的个数要不能为0
if(k == 0) {
//特殊处理K为0的情况
if((int) entry.getValue() >= 2) {
++count;
}
} else if(numMap.containsKey((int) entry.getKey() + k)) {//因为是递增的，所以可以吧相同对排除掉
++count;
}
} return count;
} public static void main(String args[]) {
int[] A = {3,1,4,1,5};
int[] B = {1,2,3,4,5};
int k = -1;
FindPairs fuc = new FindPairs();
System.out.println(fuc.solution(A, k));
}
}
