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fanxb 2021-03-30 18:35:09 +08:00
parent ac0083cb91
commit 0ff3d79282
14 changed files with 365 additions and 8 deletions
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47
5.leetcode/src/com/fanxb/common/Q122.java Normal file
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package com.fanxb.common;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
/**
* 题目地址： https://leetcode-cn.com/problems/partition-labels/
* 解题思路：首先遍历一次字符串，记录每个字母最后出现的位置
* 然后再遍历一遍，记录当前字符串的开始位置start,结束位置end. 当第i个字母的结束位置》end时end=第i个字母的结束位置，知道i=end说明当前位置为字符串的结束位置
*
* @author fanxb
* Date: 2020/6/9 15:10
*/
public class Q122 {
public static int solution(int[] prices) {
int res = 0;
int start = -1, end = -1;
for (int i = 0; i < prices.length - 1; i++) {
if (prices[i + 1] > prices[i]) {
//说明是上坡
if (start == -1) {
start = prices[i];
}
end = prices[i + 1];
} else if (prices[i + 1] < prices[i]) {
//说明开始下坡了
if (end > start) {
res += end - start;
}
start = -1;
end = -1;
}
}
if (end > start) {
res += end - start;
}
return res;
}
public static void main(String[] args) {
int[] prices = {7, 1, 5, 3, 6, 4};
System.out.println(solution(prices));
}
}

7
5.leetcode/src/com/fanxb/interview/Q1663.java → 5.leetcode/src/com/fanxb/common/Q1663.java
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@ -1,4 +1,4 @@
package com.fanxb.interview;
package com.fanxb.common;
/**
* 具有给定数值的最小字符串
@ -57,7 +57,8 @@ public class Q1663 {
}
public static void main(String[] args) {
System.out.println(new Q1663().translateNum(23100, 567226));
System.out.println(new Q1663().translateNum1(23100, 567226));
System.out.println(new Q1663().translateNum(1, 26));
System.out.println(new Q1663().translateNum1(1, 26));
System.out.println(4<<1);
}
}

29
5.leetcode/src/com/fanxb/common/Q190.java Normal file
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package com.fanxb.common;
/**
* 具有给定数值的最小字符串
* <p>
* 题目地址：https://leetcode-cn.com/problems/reverse-bits
* <p>
* 解题思路：简单的位运算
*
* @author fanxb
* Date: 2020/6/9 15:10
*/
public class Q190 {
public static int solution(int n) {
int res = 0;
for (int i = 0; i < 32; i++, n = n >>> 1) {
res = res << 1 | (n & 1);
System.out.println(Integer.toBinaryString(res));
}
return res;
}
public static void main(String[] args) {
int num = 43261596;
System.out.println(Integer.toBinaryString(num));
System.out.println(solution(num));
}
}

39
5.leetcode/src/com/fanxb/common/Q406.java Normal file
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@ -0,0 +1,39 @@
package com.fanxb.common;
import java.util.Arrays;
import java.util.LinkedList;
/**
* 题目地址：
* 解题思路：
*
*
* @author fanxb
* Date: 2020/6/9 15:10
*/
public class Q406 {
public static int[][] solution(int[][] people) {
Arrays.sort(people, (a, b) -> {
if (a[0] == b[0]) {
return Integer.compare(a[1], b[1]);
} else {
return Integer.compare(b[0], a[0]);
}
});
LinkedList<int[]> res = new LinkedList<>();
for (int[] person : people) {
if (person[1] >= res.size()) {
res.add(person);
} else {
res.add(person[1], person);
}
}
return res.toArray(new int[res.size()][]);
}
public static void main(String[] args) {
int[][] people = {{7, 0}, {4, 4}, {7, 1}, {5, 0}, {6, 1}, {5, 2}};
System.out.println(Arrays.deepToString(solution(people)));
}
}

34
5.leetcode/src/com/fanxb/common/Q435.java Normal file
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package com.fanxb.common;
import java.lang.reflect.AccessibleObject;
import java.util.Arrays;
import java.util.LinkedList;
/**
* 地址: https://leetcode-cn.com/problems/non-overlapping-intervals
*
* @author fanxb
* Date: 2020/6/11 10:58
*/
public class Q435 {
public static int solution(int[][] intervals) {
Arrays.sort(intervals, (a, b) -> a[1] - b[1]);
int count = 0, i = 0, j = 1;
while (j < intervals.length) {
if (intervals[i][1] <= intervals[j][0]) {
//说明第i个区间和第j个区间，所以将i移动到j,再和j+1比较
i = j;
} else {
//说明第i个区间和第j个区间重复,干掉j,i再和j+1比较
count++;
}
j++;
}
return count;
}
public static void main(String[] args) {
int[][] arr = {{1, 4}, {2, 3}, {1, 8}};
System.out.println(solution(arr));
}
}

36
5.leetcode/src/com/fanxb/common/Q452.java Normal file
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package com.fanxb.common;
import java.util.Arrays;
/**
* 题目地址： https://leetcode-cn.com/problems/minimum-number-of-arrows-to-burst-balloons/
* 解题思路：贪心算法，按照结束点排序，
*
* @author fanxb
* Date: 2020/6/9 15:10
*/
public class Q452 {
public static int solution(int[][] points) {
if (points.length == 0) {
return 0;
}
Arrays.sort(points, (a, b) -> Integer.compare(a[1], b[1]));
//第一根箭的位置是第一个气球的结束点
int endNum = points[0][1], count = 1;
for (int[] point : points) {
if (point[0] > endNum) {
//当某个气球的开始点>箭的位置时说明需要一根新的箭了，箭位置为当前气球的结束点
endNum = point[1];
count++;
}
}
return count;
}
public static void main(String[] args) {
int[][] s = {{-2147483646, -2147483645}, {2147483646, 2147483647}};
// int[][] s = {{10, 16}, {2, 8}, {1, 6}, {7, 12}};
System.out.println(solution(s));
}
}

2
5.leetcode/src/com/fanxb/interview/Q46.java → 5.leetcode/src/com/fanxb/common/Q46.java
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@ -1,4 +1,4 @@
package com.fanxb.interview;
package com.fanxb.common;
/**
* Created with IntelliJ IDEA

44
5.leetcode/src/com/fanxb/common/Q605.java Normal file
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package com.fanxb.common;
/**
* 具有给定数值的最小字符串
* <p>
* 题目地址：https://leetcode-cn.com/problems/smallest-string-with-a-given-numeric-value/
* <p>
* 解题思路：
* 1. 贪心算法，字典序最小意思就是要把尽量把小的字母放前面，也可以说是把尽量大的数字放后面，因此有两种做法。
* a. 每次放尽量小的字母在前面，然后判断后面的全放最大的字母能否满足（大于等于）k-已存在的数字和.如果能满足再放第二个,如果不满足就把当前的字母加大1,再判断直到满足要求
*
* @author fanxb
* Date: 2020/6/9 15:10
*/
public class Q605 {
public static boolean solution(int[] flowerbed, int n) {
//第i个位置当成花坛的起点，前一个位置肯定没种花
int i = 0;
int count = 0;
while (i < flowerbed.length) {
if (flowerbed[i] == 1) {
//第i个有花，下一个可能有花的位置肯定是i+2
i += 2;
} else {
//第i个位置没花
if (i == flowerbed.length - 1 || flowerbed[i + 1] == 0) {
//i是最后一个位置 或者 i+1没花
count++;
i += 2;
} else {
//第i个位置没花，第i+1个位置有花，说明下一个种花的起点至少是i+3
i += 3;
}
}
}
return count >= n;
}
public static void main(String[] args) {
int[] s = {1, 0, 0, 0, 0, 1};
System.out.println(solution(s, 2));
}
}

2
5.leetcode/src/com/fanxb/interview/Q61.java → 5.leetcode/src/com/fanxb/common/Q61.java
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@ -1,4 +1,4 @@
package com.fanxb.interview;
package com.fanxb.common;
/**
* 旋转链表

2
5.leetcode/src/com/fanxb/interview/Q64.java → 5.leetcode/src/com/fanxb/common/Q64.java
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@ -1,4 +1,4 @@
package com.fanxb.interview;
package com.fanxb.common;
/**
* Created with IntelliJ IDEA

52
5.leetcode/src/com/fanxb/common/Q665.java Normal file
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@ -0,0 +1,52 @@
package com.fanxb.common;
import java.util.Arrays;
import java.util.LinkedList;
/**
* 题目地址：https://leetcode-cn.com/problems/non-decreasing-array/submissions/
* 解题思路：贪心算法，遍历数组，每次调整都要尽量用最小的值
*
* @author fanxb
* Date: 2020/6/9 15:10
*/
public class Q665 {
public static boolean betterSolution(int[] nums) {
if (nums.length <= 2) {
return true;
}
//是否首次调整
boolean firstAdjust = true;
for (int i = 0; i < nums.length - 1; i++) {
if (nums[i] > nums[i + 1]) {
if (i == 0) {
//首个节点就不满足只能让nums[0]=nums[1]一种调整办法
firstAdjust = false;
nums[0] = nums[1];
} else {
//不为首个节点
if (!firstAdjust) {
return false;
}
if (nums[i + 1] < nums[i - 1]) {
//如果i+1的值小于i-1的值只能将i+1的值调大
nums[i + 1] = nums[i];
} else {
//否则将nums[i]调小是最好的
nums[i] = nums[i - 1];
}
firstAdjust = false;
}
}
}
return true;
}
public static void main(String[] args) {
int[] nums = {-1, 4, 2, 3};
System.out.println(betterSolution(nums));
}
}

34
5.leetcode/src/com/fanxb/common/Q74.java Normal file
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@ -0,0 +1,34 @@
package com.fanxb.common;
import java.util.Arrays;
/**
* 具有给定数值的最小字符串
* 题目地址： https://leetcode-cn.com/problems/search-a-2d-matrix/
* 解题思路：两次二分即可，第一次二分列找到所在行，然后二分所在行找到是否存在目标值（由于行列数少于100,直接搜索问题也不大）
*
*
* @author fanxb
* Date: 2020/6/9 15:10
*/
public class Q74 {
public static boolean solution(int[][] matrix, int target) {
//m行数，n列数
int m = matrix.length, n = matrix[0].length;
if (matrix[0][0] > target || matrix[m - 1][n - 1] < target) {
return false;
}
for (int i = 1; i < m; i++) {
if (matrix[i][0] > target) {
return Arrays.stream(matrix[i - 1]).anyMatch(item -> item == target);
}
}
return Arrays.stream(matrix[m - 1]).anyMatch(item -> item == target);
}
public static void main(String[] args) {
int[][] s = {{1, 3, 5, 7}, {10, 11, 16, 20}, {23, 30, 34, 6}};
System.out.println(solution(s, 3));
}
}

41
5.leetcode/src/com/fanxb/common/Q763.java Normal file
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@ -0,0 +1,41 @@
package com.fanxb.common;
import java.util.*;
/**
* 题目地址： https://leetcode-cn.com/problems/partition-labels/
* 解题思路：首先遍历一次字符串，记录每个字母最后出现的位置
* 然后再遍历一遍，记录当前字符串的开始位置start,结束位置end. 当第i个字母的结束位置》end时end=第i个字母的结束位置，知道i=end说明当前位置为字符串的结束位置
*
* @author fanxb
* Date: 2020/6/9 15:10
*/
public class Q763 {
public static List<Integer> solution(String S) {
List<Integer> res = new LinkedList<>();
Map<Character, Integer> map = new HashMap<>(24);
char[] chars = new char[S.length()];
for (int i = 0; i < chars.length; i++) {
chars[i] = S.charAt(i);
map.put(chars[i], i);
}
int start = 0, end = map.get(chars[0]), temp;
for (int i = 0; i < chars.length; i++) {
if ((temp = map.get(chars[i])) > end) {
//中间的某个点最远出现的位置大于end,说明当前字符串的结束点至少为end
end = temp;
} else if (i == end) {
res.add(end - start + 1);
//进行下一段的搜索
start = end + 1;
}
}
return res;
}
public static void main(String[] args) {
String S = "eaaaabaaec";
System.out.println(solution(S));
}
}

4
5.leetcode/src/com/fanxb/interview/Q83.java → 5.leetcode/src/com/fanxb/common/Q83.java
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@ -1,4 +1,4 @@
package com.fanxb.interview;
package com.fanxb.common;
/**
* TODO 类描述
@ -41,7 +41,7 @@ public class Q83 {
* 递归版块
*
* @param head head
* @return com.fanxb.interview.Q83.ListNode
* @return com.fanxb.common.Q83.ListNode
* @author fanxb
* @date 2021/3/26
**/