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Commit 655ede2

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feat: add solutions to lcp problem: No.61
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‎lcp/LCP 61. 气温变化趋势/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:动态规划**
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我们用变量 $f$ 维护当前趋势相同的连续天数,用变量 $ans$ 维护最大的连续天数。
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遍历数组,对于第 $i$ 天,记两地的气温变化趋势分别为 $x$ 和 $y,ドル如果 $x$ 和 $y$ 均为 0ドル$ 或者 $x$ 和 $y$ 均为正数或负数,则说明第 $i$ 天和第 $i+1$ 天的气温变化趋势相同,此时 $f$ 自增 1ドル,ドル并更新 $ans$;否则说明第 $i$ 天和第 $i+1$ 天的气温变化趋势不同,此时 $f$ 重置为 0ドル$。
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最终返回 $ans$ 即可。
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时间复杂度 $O(n),ドル空间复杂度 $O(1)$。其中 $n$ 为数组长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def temperatureTrend(self, temperatureA: List[int], temperatureB: List[int]) -> int:
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ans = f = 0
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n = len(temperatureA)
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for i in range(n - 1):
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x = temperatureA[i + 1] - temperatureA[i]
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y = temperatureB[i + 1] - temperatureB[i]
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if x == y == 0 or x * y > 0:
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f += 1
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ans = max(ans, f)
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else:
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f = 0
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int temperatureTrend(int[] temperatureA, int[] temperatureB) {
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int ans = 0, f = 0;
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for (int i = 0; i < temperatureA.length - 1; ++i) {
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int x = temperatureA[i + 1] - temperatureA[i];
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int y = temperatureB[i + 1] - temperatureB[i];
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if ((x == 0 && y == 0) || x * y > 0) {
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ans = Math.max(ans, ++f);
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} else {
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f = 0;
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int temperatureTrend(vector<int>& temperatureA, vector<int>& temperatureB) {
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int ans = 0, f = 0;
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for (int i = 0; i < temperatureA.size() - 1; ++i) {
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int x = temperatureA[i + 1] - temperatureA[i];
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int y = temperatureB[i + 1] - temperatureB[i];
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if ((x == 0 && y == 0) || x * y > 0) {
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ans = max(ans, ++f);
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} else {
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f = 0;
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func temperatureTrend(temperatureA []int, temperatureB []int) int {
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ans, f := 0, 0
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for i := range temperatureA[1:] {
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x := temperatureA[i+1] - temperatureA[i]
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y := temperatureB[i+1] - temperatureB[i]
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if (x == 0 && y == 0) || x*y > 0 {
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f++
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ans = max(ans, f)
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} else {
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f = 0
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**
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class Solution {
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public:
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int temperatureTrend(vector<int>& temperatureA, vector<int>& temperatureB) {
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int ans = 0, f = 0;
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for (int i = 0; i < temperatureA.size() - 1; ++i) {
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int x = temperatureA[i + 1] - temperatureA[i];
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int y = temperatureB[i + 1] - temperatureB[i];
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if ((x == 0 && y == 0) || x * y > 0) {
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ans = max(ans, ++f);
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} else {
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f = 0;
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}
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}
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return ans;
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}
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};
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func temperatureTrend(temperatureA []int, temperatureB []int) int {
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ans, f := 0, 0
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for i := range temperatureA[1:] {
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x := temperatureA[i+1] - temperatureA[i]
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y := temperatureB[i+1] - temperatureB[i]
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if (x == 0 && y == 0) || x*y > 0 {
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f++
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ans = max(ans, f)
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} else {
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f = 0
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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class Solution {
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public int temperatureTrend(int[] temperatureA, int[] temperatureB) {
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int ans = 0, f = 0;
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for (int i = 0; i < temperatureA.length - 1; ++i) {
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int x = temperatureA[i + 1] - temperatureA[i];
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int y = temperatureB[i + 1] - temperatureB[i];
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if ((x == 0 && y == 0) || x * y > 0) {
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ans = Math.max(ans, ++f);
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} else {
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f = 0;
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}
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}
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return ans;
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}
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}
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class Solution:
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def temperatureTrend(self, temperatureA: List[int], temperatureB: List[int]) -> int:
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ans = f = 0
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n = len(temperatureA)
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for i in range(n - 1):
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x = temperatureA[i + 1] - temperatureA[i]
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y = temperatureB[i + 1] - temperatureB[i]
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if x == y == 0 or x * y > 0:
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f += 1
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ans = max(ans, f)
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else:
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f = 0
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return ans

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