博主的解法过于冗长，是一直对着不同的案例debug修改出来的，不建议学习。虽然提交成功了，但是自己最后都不知道写的是啥了哈哈哈。
package keepcoding.leetcode.leetcode2409;
/*Alice 和 Bob 计划分别去罗马开会。给你四个字符串 arriveAlice ，leaveAlice ，arriveBob 和 leaveBob 。Alice 会在日期 arriveAlice 到 leaveAlice 之间在城市里（日期为闭区间），而 Bob 在日期 arriveBob 到 leaveBob 之间在城市里（日期为闭区间）。每个字符串都包含 5 个字符，格式为 "MM-DD" ，对应着一个日期的月和日。请你返回 Alice和 Bob 同时在罗马的天数。你可以假设所有日期都在 同一个 自然年，而且 不是 闰年。每个月份的天数分别为：[31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]*/
public class Result01 {public static void main(String[] args) {int days = countDaysTogether("08-12","08-26","08-26","10-25");System.out.println(days);}/*输入：arriveAlice = "08-15", leaveAlice = "08-18",arriveBob = "08-16", leaveBob = "08-19"输出：3解释：Alice 从 8 月 15 号到 8 月 18 号在罗马。Bob 从 8 月 16 号到 8 月 19 号在罗马，他们同时在罗马的日期为 8 月 16、17 和 18 号。所以答案为 3 。*/public static int countDaysTogether(String arriveAlice, String leaveAlice, String arriveBob, String leaveBob) {String[] arriveAliceTimeArray = arriveAlice.split("-",2);String[] leaveAliceTimeArray = leaveAlice.split("-",2);String[] arriveBobTimeArray = arriveBob.split("-",2);String[] leaveBobTimeArray = leaveBob.split("-",2);int aliceArriveMonth = Integer.valueOf(arriveAliceTimeArray[0]);int aliceLeaveMonth = Integer.valueOf(leaveAliceTimeArray[0]);int aliceArriveDay = Integer.valueOf(arriveAliceTimeArray[1]);int aliceLeaveDay = Integer.valueOf(leaveAliceTimeArray[1]);int bobArriveMonth = Integer.valueOf(arriveBobTimeArray[0]);int bobLeaveMonth = Integer.valueOf(leaveBobTimeArray[0]);int bobArriveDay = Integer.valueOf(arriveBobTimeArray[1]);int bobLeaveDay = Integer.valueOf(leaveBobTimeArray[1]);if (aliceArriveDay==bobArriveDay && aliceArriveMonth==bobArriveMonth && aliceLeaveDay==bobLeaveDay && aliceLeaveMonth==bobLeaveMonth){if (aliceArriveDay==1 && aliceLeaveMonth==12){return 365;}else if (aliceArriveDay==28 && aliceArriveMonth==2 && aliceLeaveMonth==3){return 2;}}//alice和bob同一个月到if (aliceArriveMonth == bobArriveMonth) {//同一个月走if (aliceLeaveMonth==bobLeaveMonth){if (aliceArriveDay > bobLeaveDay || bobArriveDay > aliceLeaveDay) {//alice/bob到的时候 bob/alice已经走了 共同度过的日子为0return 0;} else {//alice先到if (aliceArriveDay <= bobArriveDay) {return aliceLeaveDay <= bobLeaveDay ? (aliceLeaveDay-bobArriveDay)+1 : (bobLeaveDay-bobArriveDay)+1;//alice后到} else {return aliceLeaveDay <= bobLeaveDay ? (aliceLeaveDay-aliceArriveDay)+1 : (bobLeaveDay-aliceArriveDay)+1;}}//alice走的月份小于bob 即alice先走}else if (aliceLeaveMonth<bobLeaveMonth){if (aliceLeaveDay==bobArriveDay){return 1;}else {int sumDays = aliceLeaveDay;for (int i = aliceArriveMonth+1; i < aliceLeaveMonth; i++) {sumDays += getMonthDays(i);}sumDays += aliceArriveDay>=bobArriveDay ? getMonthDays(aliceArriveMonth)-aliceArriveDay : getMonthDays(bobArriveMonth)-bobArriveDay;return sumDays+1;}//alice走的月份大于bob 即bob先走}else {int sumDays = 0;if (bobArriveMonth==bobLeaveMonth){sumDays += bobLeaveDay - bobArriveDay;}else {sumDays = bobLeaveDay;for (int i = bobArriveMonth+1; i <bobLeaveMonth ; i++) {sumDays += getMonthDays(i);}sumDays += aliceArriveDay>=bobArriveDay ? getMonthDays(aliceArriveMonth)-aliceArriveDay : getMonthDays(bobArriveMonth)-bobArriveDay;}return sumDays+1;}//alice月份先到} else if (aliceArriveMonth < bobArriveMonth) {if (aliceLeaveMonth < bobArriveMonth) {//alice在bob到之前就走了return 0;//alice 走 的那个月 bob 到} else if (aliceLeaveMonth == bobArriveMonth) {if(aliceLeaveDay<=bobLeaveDay){return bobArriveDay < aliceLeaveDay ? (aliceLeaveDay-bobArriveDay)+1 : 0;}else {return bobArriveDay < aliceLeaveDay ? (bobLeaveDay-bobArriveDay)+1 : 0;}//alice在bob到的月份之后的月份才走} else {//bob走的月份小于aliceif (bobLeaveMonth < aliceLeaveMonth) {int sumDays = getMonthDays(bobArriveMonth) - bobArriveDay;for (int i = bobArriveMonth + 1; i < bobLeaveMonth; i++) {sumDays += getMonthDays(i);}sumDays += bobLeaveDay;return sumDays+1;//bob走的月份大于alice} else if (bobLeaveMonth > aliceLeaveMonth) {int sumDays = getMonthDays(bobArriveMonth) - bobArriveDay;for (int i = bobArriveMonth + 1; i < aliceLeaveMonth; i++) {sumDays += getMonthDays(i);}sumDays += aliceLeaveDay;return sumDays+1;//bob跟alice一个月走}else {int sumDays = getMonthDays(bobArriveMonth) - bobArriveDay;for (int i = bobArriveMonth + 1; i < aliceLeaveMonth; i++) {sumDays += getMonthDays(i);}sumDays += Math.min(aliceLeaveDay,bobLeaveDay);return sumDays+1;}}//bob月份先到} else {if (bobLeaveMonth < aliceArriveMonth) {//bob在alice到之前就走了return 0;//bob 走 的那个月 alice 到} else if (bobLeaveMonth == aliceArriveMonth) {return aliceArriveDay < bobLeaveDay ? (bobLeaveDay-aliceArriveDay)+1 : 0;//bob在alice到的月份之后的月份才走} else {//alice走的月份小于bobif (aliceLeaveMonth < bobLeaveMonth) {int sumDays = getMonthDays(aliceArriveMonth) - aliceArriveDay;for (int i = aliceArriveMonth + 1; i < aliceLeaveMonth; i++) {sumDays += getMonthDays(i);}sumDays += aliceLeaveDay;return sumDays+1;//alice走的月份大于bob} else if (aliceLeaveMonth > bobLeaveMonth) {int sumDays = getMonthDays(aliceArriveMonth) - aliceArriveDay;for (int i = aliceArriveMonth + 1; i < bobLeaveMonth; i++) {sumDays += getMonthDays(i);}sumDays += bobLeaveDay;return sumDays+1;//alice跟bob一个月走}else {int sumDays = getMonthDays(aliceArriveMonth) - aliceArriveDay;for (int i = aliceArriveMonth + 1; i < aliceLeaveMonth; i++) {sumDays += getMonthDays(i);}sumDays += Math.min(aliceLeaveDay,bobLeaveDay);return sumDays+1;}}}}//获取月份对应的天数public static int getMonthDays(int i){switch (i){case 1: return 31;case 2: return 28;case 3: return 31;case 4: return 30;case 5: return 31;case 6: return 30;case 7: return 31;case 8: return 31;case 9: return 30;case 10: return 31;case 11: return 30;case 12: return 31;default: return 0;}}
}官方解较为巧妙，大家可以学习：
package keepcoding.leetcode.leetcode2409;
/*
我们可以设计一个函数 calculateDayOfYear 来计算输入中的每个日子在一年中是第几天。
计算输入中的每个日子在一年中是第几天时，可以利用前缀和数组来降低每次计算的复杂度。
知道每个日子是一年中的第几天后，可以先通过比较算出两人到达日子的最大值，离开日子的最小值，然后利用减法计算重合的日子
*/public class Result02 {public static void main(String[] args) {int days = countDaysTogether("10-01","11-01","11-01","12-31");System.out.println(days);}public static int countDaysTogether(String arriveAlice, String leaveAlice, String arriveBob, String leaveBob) {//将每个月份的天数存入数组int[] datesOfMonths = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};//累加数组——计算每个月在这一年已经过了多少天int[] prefixSum = new int[13];for (int i = 0; i < 12; i++) {//i+1 是因为要在数组1-12的位置上一一对应的存入月份累加的天数 eg.一月这一年过了31天 二月过了31+28 三月31+28+31 ...prefixSum[i + 1] = prefixSum[i] + datesOfMonths[i];}//计算alice到的天数是一年的第几天int arriveAliceDay = calculateDayOfYear(arriveAlice, prefixSum);//计算alice走的天数是一年的第几天int leaveAliceDay = calculateDayOfYear(leaveAlice, prefixSum);//计算bob到的天数是一年的第几天int arriveBobDay = calculateDayOfYear(arriveBob, prefixSum);//计算bob走的天数是一年的第几天int leaveBobDay = calculateDayOfYear(leaveBob, prefixSum);//用Math.min(leaveAliceDay, leaveBobDay)计算谁先走的，Math.max(arriveAliceDay, arriveBobDay)计算谁后到的//如果存在共处的时间，先走的-后到的即为共处的天数//如果先走的-后到的 < 0 证明没有相遇共处的时间——return 0；// +1 是因为 如果alice走的那天bob到 根据题意这也算共处了一天 eg .alice 10-01到 10-31走 bob 10-31到 11-3走 ;根据先走的-后到的计算=0，但是有共处的一天，所以+1return Math.max(0, Math.min(leaveAliceDay, leaveBobDay) - Math.max(arriveAliceDay, arriveBobDay) + 1);}//计算是哪一天到的public static int calculateDayOfYear(String day, int[] prefixSum) {//String类的substring()方法 ——截取字符串 https://blog.csdn.net/Crezfikbd/article/details/119708978int month = Integer.parseInt(day.substring(0, 2));int date = Integer.parseInt(day.substring(3));return prefixSum[month - 1] + date;//到的这个月的前一个月总共过了多少天 再加上这个月到的日期 即为这一年的第几天}
}知道思路后自己又手敲了一遍：
package keepcoding.leetcode.leetcode2409;public class DoItAgain {public static void main(String[] args) {int days = countTogetherDays("10-01","11-01","11-01","12-31");System.out.println(days);}//计算哪天到public static int countTogetherDays(String aliceArrive,String aliceLeave,String bobArrive,String bobLeave){int[] monthDays = {31,28,31,30,31,30,31,31,30,31,30,31};int[] preMonthSum = new int[13];for (int i = 0; i < monthDays.length; i++) {preMonthSum[i+1] = preMonthSum[i] + monthDays[i];}//计算alice一年中的第几天到int aComDay = countDay(aliceArrive,preMonthSum);//计算alice一年中的第几天走int aGoDay = countDay(aliceLeave,preMonthSum);//计算bob一年中的第几天到int bComDay = countDay(bobArrive,preMonthSum);//计算bob一年中的第几天走int bGoDay = countDay(bobLeave,preMonthSum);if (bComDay>aGoDay || aComDay>bGoDay){//没有相遇return 0;}else {//先走的-后到的日期return Math.min(aGoDay,bGoDay) - Math.max(aComDay,bComDay) + 1;}}//计算各个时间点是这一年的第多少天public static int countDay(String s,int[] preMonthSum){//转化截取的字符串——得到到达的月份、日期int month = Integer.parseInt(s.substring(0,2));int day = Integer.parseInt(s.substring(3));//根据累加的数组结合当月到达的日期 计算到达的天数return preMonthSum[month-1]+day;}
}
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