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| 1 | +# [2409. 统计共同度过的日子数](https://leetcode.cn/problems/count-days-spent-together) |
| 2 | + |
| 3 | +[English Version](/solution/2400-2499/2409.Count%20Days%20Spent%20Together/README_EN.md) |
| 4 | + |
| 5 | +## 题目描述 |
| 6 | + |
| 7 | +<!-- 这里写题目描述 --> |
| 8 | + |
| 9 | +<p>Alice 和 Bob 计划分别去罗马开会。</p> |
| 10 | + |
| 11 | +<p>给你四个字符串 <code>arriveAlice</code> ，<code>leaveAlice</code> ，<code>arriveBob</code> 和 <code>leaveBob</code> 。Alice 会在日期 <code>arriveAlice</code> 到 <code>leaveAlice</code> 之间在城市里（<strong>日期为闭区间</strong>），而 Bob 在日期 <code>arriveBob</code> 到 <code>leaveBob</code> 之间在城市里（<strong>日期为闭区间</strong>）。每个字符串都包含 5 个字符，格式为 <code>"MM-DD"</code> ，对应着一个日期的月和日。</p> |
| 12 | + |
| 13 | +<p>请你返回 Alice和 Bob 同时在罗马的天数。</p> |
| 14 | + |
| 15 | +<p>你可以假设所有日期都在 <strong>同一个</strong> 自然年，而且 <strong>不是</strong> 闰年。每个月份的天数分别为：<code>[31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]</code> 。</p> |
| 16 | + |
| 17 | +<p> </p> |
| 18 | + |
| 19 | +<p><strong>示例 1：</strong></p> |
| 20 | + |
| 21 | +<pre> |
| 22 | +<b>输入：</b>arriveAlice = "08-15", leaveAlice = "08-18", arriveBob = "08-16", leaveBob = "08-19" |
| 23 | +<b>输出：</b>3 |
| 24 | +<b>解释：</b>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 。 |
| 25 | +</pre> |
| 26 | + |
| 27 | +<p><strong>示例 2：</strong></p> |
| 28 | + |
| 29 | +<pre> |
| 30 | +<b>输入：</b>arriveAlice = "10-01", leaveAlice = "10-31", arriveBob = "11-01", leaveBob = "12-31" |
| 31 | +<b>输出：</b>0 |
| 32 | +<b>解释：</b>Alice 和 Bob 没有同时在罗马的日子，所以我们返回 0 。 |
| 33 | +</pre> |
| 34 | + |
| 35 | +<p> </p> |
| 36 | + |
| 37 | +<p><strong>提示：</strong></p> |
| 38 | + |
| 39 | +<ul> |
| 40 | + <li>所有日期的格式均为 <code>"MM-DD"</code> 。</li> |
| 41 | + <li>Alice 和 Bob 的到达日期都 <strong>早于或等于</strong> 他们的离开日期。</li> |
| 42 | + <li>题目测试用例所给出的日期均为 <strong>非闰年</strong> 的有效日期。</li> |
| 43 | +</ul> |
| 44 | + |
| 45 | +## 解法 |
| 46 | + |
| 47 | +<!-- 这里可写通用的实现逻辑 --> |
| 48 | + |
| 49 | +**方法一：模拟** |
| 50 | + |
| 51 | +将日期转换为天数，然后计算两个人在罗马的天数。 |
| 52 | + |
| 53 | +时间复杂度 $O(1)$。 |
| 54 | + |
| 55 | +<!-- tabs:start --> |
| 56 | + |
| 57 | +### **Python3** |
| 58 | + |
| 59 | +<!-- 这里可写当前语言的特殊实现逻辑 --> |
| 60 | + |
| 61 | +```python |
| 62 | +class Solution: |
| 63 | + def countDaysTogether(self, arriveAlice: str, leaveAlice: str, arriveBob: str, leaveBob: str) -> int: |
| 64 | + if leaveAlice < arriveBob or leaveBob < arriveAlice: |
| 65 | + return 0 |
| 66 | + a = max(arriveAlice, arriveBob) |
| 67 | + b = min(leaveAlice, leaveBob) |
| 68 | + days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] |
| 69 | + x = sum(days[:int(a[:2]) - 1]) + int(a[3:]) |
| 70 | + y = sum(days[:int(b[:2]) - 1]) + int(b[3:]) |
| 71 | + return y - x + 1 |
| 72 | +``` |
| 73 | + |
| 74 | +### **Java** |
| 75 | + |
| 76 | +<!-- 这里可写当前语言的特殊实现逻辑 --> |
| 77 | + |
| 78 | +```java |
| 79 | +class Solution { |
| 80 | + private int[] days = new int[] {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; |
| 81 | + |
| 82 | + public int countDaysTogether(String arriveAlice, String leaveAlice, String arriveBob, String leaveBob) { |
| 83 | + String a = arriveAlice.compareTo(arriveBob) < 0 ? arriveBob : arriveAlice; |
| 84 | + String b = leaveAlice.compareTo(leaveBob) < 0 ? leaveAlice : leaveBob; |
| 85 | + int x = f(a), y = f(b); |
| 86 | + return Math.max(y - x + 1, 0); |
| 87 | + } |
| 88 | + |
| 89 | + private int f(String s) { |
| 90 | + int i = Integer.parseInt(s.substring(0, 2)) - 1; |
| 91 | + int res = 0; |
| 92 | + for (int j = 0; j < i; ++j) { |
| 93 | + res += days[j]; |
| 94 | + } |
| 95 | + res += Integer.parseInt(s.substring(3)); |
| 96 | + return res; |
| 97 | + } |
| 98 | +} |
| 99 | +``` |
| 100 | + |
| 101 | +### **C++** |
| 102 | + |
| 103 | +```cpp |
| 104 | +class Solution { |
| 105 | +public: |
| 106 | + vector<int> days = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; |
| 107 | + |
| 108 | + int countDaysTogether(string arriveAlice, string leaveAlice, string arriveBob, string leaveBob) { |
| 109 | + string a = arriveAlice < arriveBob ? arriveBob : arriveAlice; |
| 110 | + string b = leaveAlice < leaveBob ? leaveAlice : leaveBob; |
| 111 | + int x = f(a), y = f(b); |
| 112 | + return max(0, y - x + 1); |
| 113 | + } |
| 114 | + |
| 115 | + int f(string s) { |
| 116 | + int m, d; |
| 117 | + sscanf(s.c_str(), "%d-%d", &m, &d); |
| 118 | + int res = 0; |
| 119 | + for (int i = 0; i < m - 1; ++i) { |
| 120 | + res += days[i]; |
| 121 | + } |
| 122 | + res += d; |
| 123 | + return res; |
| 124 | + } |
| 125 | +}; |
| 126 | +``` |
| 127 | +
|
| 128 | +### **Go** |
| 129 | +
|
| 130 | +```go |
| 131 | +func countDaysTogether(arriveAlice string, leaveAlice string, arriveBob string, leaveBob string) int { |
| 132 | + days := []int{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31} |
| 133 | + f := func(s string) int { |
| 134 | + m, _ := strconv.Atoi(s[:2]) |
| 135 | + d, _ := strconv.Atoi(s[3:]) |
| 136 | + res := 0 |
| 137 | + for i := 0; i < m-1; i++ { |
| 138 | + res += days[i] |
| 139 | + } |
| 140 | + res += d |
| 141 | + return res |
| 142 | + } |
| 143 | + a, b := arriveAlice, leaveBob |
| 144 | + if arriveAlice < arriveBob { |
| 145 | + a = arriveBob |
| 146 | + } |
| 147 | + if leaveAlice < leaveBob { |
| 148 | + b = leaveAlice |
| 149 | + } |
| 150 | + x, y := f(a), f(b) |
| 151 | + ans := y - x + 1 |
| 152 | + if ans < 0 { |
| 153 | + return 0 |
| 154 | + } |
| 155 | + return ans |
| 156 | +} |
| 157 | +``` |
| 158 | + |
| 159 | +### **TypeScript** |
| 160 | + |
| 161 | +```ts |
| 162 | + |
| 163 | +``` |
| 164 | + |
| 165 | +### **...** |
| 166 | + |
| 167 | +``` |
| 168 | +
|
| 169 | +
|
| 170 | +``` |
| 171 | + |
| 172 | +<!-- tabs:end --> |
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