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2017-06-11 （4）
2017-06-12 （3）
2017-06-13 （5）
2017-06-14 （2）
2017-06-15 （2）
2017-06-17 （7）
2017-06-18 （0）：上午去面试了 下午去室友家玩了。有点放纵。
2017-06-19 （3）
2017-06-20 （5）:学到了求不同组合子集，相加等于k的组合等问题的通用接发backtrack。;反转数组（3步 全部反转 反转序列1 反转序列2）
2017-06-21（4）主要是二分法
2017-06-22（2）学习了kadana算法 主要是解决连续最大子序列 和最小子序列等问题
Array Nesting(565)
A zero-indexed array A consisting of N different integers is given. The array contains all integers in the range [0, N - 1].
Sets S[K] for 0 <= K < N are defined as follows:
S[K] = { A[K], A[A[K]], A[A[A[K]]], ... }.
Sets S[K] are finite for each K and should NOT contain duplicates.
Write a function that given an array A consisting of N integers, return the size of the largest set S[K] for this array.
Example 1:
Input: A = [5,4,0,3,1,6,2]
Output: 4
Explanation:
A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.
One of the longest S[K]:
S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}
Note:
- N is an integer within the range [1, 20,000].
- The elements of A are all distinct.
- Each element of array A is an integer within the range [0, N-1].
思路
遍历一次数组，记录每次的count，找到最大值。其中找count的时候，可以用一个数组进行标记下，是否该值访问过。
疑问？ 每次不是应该把flag重置么
import java.util.Arrays;
public class Solution {
public int arrayNesting(int[] nums) {
int count=0;
int max=0;
int[] flag=new int[nums.length];
for(int i=0;icount?max:count;
}
return max;
}
public int arrayNesting(int[] a) {
int maxsize = 0;
for (int i = 0; i < a.length; i++) {
int size = 0;
for (int k = i; a[k] >= 0; size++) {
int ak = a[k];
a[k] = -1; // mark a[k] as visited;
k = ak;
}
maxsize = Integer.max(maxsize, size);
}
return maxsize;
}
}
Array Partition I(561)
Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.
Example 1:
Input: [1,4,3,2]
Output: 4
Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).
Note:
- n is a positive integer, which is in the range of [1, 10000].
- All the integers in the array will be in the range of [-10000, 10000].
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public class Solution {
public int arrayPairSum(int[] nums) {
Arrays.sort(nums);
int sum=0;
for(int i=0;iCan Place Flowers(605)
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 nnew flowers can be planted in it without violating the no-adjacent-flowers rule.
Example 1:
Input: flowerbed = [1,0,0,0,1], n = 1
Output: True
Example 2:
Input: flowerbed = [1,0,0,0,1], n = 2
Output: False
Note:
- The input array won't violate no-adjacent-flowers rule.
- The input array size is in the range of [1, 20000].
- n is a non-negative integer which won't exceed the input array size.
思路
对于每一个空格 检测前面和后面是不是0 是的话 自己设为1 count+1；
package leetcode_array;
public class Can_Place_Flowers {
public boolean canPlaceFlowers(int[] flowerbed, int n) {
int count = 0;
for(int i = 0; i < flowerbed.length && count < n; i++) {
if(flowerbed[i] == 0) {
//get next and prev flower bed slot values. If i lies at the ends the next and prev are considered as 0.
int next = (i == flowerbed.length - 1) ? 0 : flowerbed[i + 1];
int prev = (i == 0) ? 0 : flowerbed[i - 1];
if(next == 0 && prev == 0) {
flowerbed[i] = 1;
count++;
}
}
}
return count == n;
}
}
Reshape the Matrix(566)
n MATLAB, there is a very useful function called 'reshape', which can reshape a matrix into a new one with different size but keep its original data.
You're given a matrix represented by a two-dimensional array, and two positive integers r and c representing the row number and columnnumber of the wanted reshaped matrix, respectively.
The reshaped matrix need to be filled with all the elements of the original matrix in the same row-traversing order as they were.
If the 'reshape' operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.
Example 1:
Input:
nums =
[[1,2],
[3,4]]
r = 1, c = 4
Output:
[[1,2,3,4]]
Explanation:
The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.
Example 2:
Input:
nums =
[[1,2],
[3,4]]
r = 2, c = 4
Output:
[[1,2],
[3,4]]
Explanation:
There is no way to reshape a 2 * 2 matrix to a 2 * 4 matrix. So output the original matrix.
Note:
- The height and width of the given matrix is in range [1, 100].
- The given r and c are all positive.
public class Solution {
public int[][] matrixReshape(int[][] nums, int r, int c) {
int n = nums.length, m = nums[0].length;
if (r*c != n*m) return nums;
int[][] res = new int[r][c];
for (int i=0;iShortest Unsorted Continuous Subarray(581)
Given an integer array, you need to find one continuous subarray that if you only sort this subarray in ascending order, then the whole array will be sorted in ascending order, too.
You need to find the shortest such subarray and output its length.
Example 1:
Input: [2, 6, 4, 8, 10, 9, 15]
Output: 5
Explanation: You need to sort [6, 4, 8, 10, 9] in ascending order to make the whole array sorted in ascending order.
Note:
- Then length of the input array is in range [1, 10,000].
- The input array may contain duplicates, so ascending order here means <=.
package leetcode_array;
import java.util.Arrays;
public class Shortest_Unsorted_Continuous_Subarray {
public int findUnsortedSubarray(int[] nums) {
int n = nums.length;
int[] temp = new int[n];
for (int i = 0; i < n; i++) temp[i] = nums[i];
Arrays.sort(temp);
int start = 0;
while (start < n && nums[start] == temp[start]) start++;
int end = n - 1;
while (end > start && nums[end] == temp[end]) end--;
return end - start + 1;
}
}
Subarray Sum Equals K(560)
Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.
Example 1:
Input:nums = [1,1,1], k = 2
Output: 2
Note:
- The length of the array is in range [1, 20,000].
- The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].
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Solution 1. Brute force. We just need two loops (i, j) and test if SUM[i, j] = k. Time complexity O(n^2), Space complexity O(1). I bet this solution will TLE.
但是暴力也能过。。
Solution 2. From solution 1, we know the key to solve this problem is SUM[i, j]. So if we know SUM[0, i - 1] and SUM[0, j], then we can easily get SUM[i, j]. To achieve this, we just need to go through the array, calculate the current sum and save number of all seen PreSum to a HashMap. Time complexity O(n), Space complexity O(n).
我们已经知道sum（i，j）=k。然后每次我们只需要来判断 和为sum-k的，存在么。存在的话，加一，不存在把当前的值，放进sum中。
public class Solution {
public int subarraySum(int[] nums, int k) {
int sum = 0, result = 0;
Map preSum = new HashMap<>();
preSum.put(0, 1);
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
if (preSum.containsKey(sum - k)) {
result += preSum.get(sum - k);
}
preSum.put(sum, preSum.getOrDefault(sum, 0) + 1);
}
return result;
}
}
public class Solution {
public int subarraySum(int[] nums, int k) {
int count=0;
for(int i=0;i Valid Triangle Number(611)
Given an array consists of non-negative integers, your task is to count the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle.
Example 1:
Input: [2,2,3,4]
Output: 3
Explanation:
Valid combinations are:
2,3,4 (using the first 2)
2,3,4 (using the second 2)
2,2,3
Note:
- The length of the given array won't exceed 1000.
- The integers in the given array are in the range of [0, 1000].
思路
先排序，然后枚举三角形最小的两条边x和y。对于第三条边z：利用三角形两边之和小于第三边的性质，即z
package leetcode_array;
import java.util.Arrays;
public class Valid_Triangle_Number {
public static int triangleNumber(int[] nums) {
Arrays.sort(nums);
int left,right,mid;
int ans = 0;
int n=nums.length;
if(n<3) return 0;
for(int i=0;i=z)
right=mid;
else
left=mid;
}
ans+=left-j;
}
}
return ans;
}
public static void main(String[] args) {
int[] arr={2,2,3,4};
System.out.println(triangleNumber(arr));
}
}