标签：app 个数 ++ HERE info find write 否则 line
Give an integer array，find the longest increasing continuous subsequence in this array.
An increasing continuous subsequence:
O(n) time and O(1) extra space.
For [5, 4, 2, 1, 3], the LICS is [5, 4, 2, 1], return 4.
For [5, 1, 2, 3, 4], the LICS is [1, 2, 3, 4], return 4.
解题：记录连续增大或者连续减少的个数，返回最大值。代码如下：
public class Solution {
/**
* @param A: An array of Integer
* @return: an integer
*/
public int longestIncreasingContinuousSubsequence(int[] A) {
// write your code here
int i_count = 1;//上升的时候的个数
int d_count = 1;//下降时候的个数
int temp = 1;
//注意，当下标有减号时，要注意返回，下标不为负
if(A.length == 0){
return 0;
}
for(int i = 1; i < A.length; i++){
if(A[i] > A[i-1]){
//说明增大
temp++;
}else{
//否则
if(temp > i_count){
i_count = temp;//更新
}
temp = 1;
}
}
if(temp > i_count){
//如果一直到最后，可能缺少一次跟新
i_count = temp;
}
temp = 1;
for(int i = 1; i < A.length; i++){
if(A[i] < A[i-1]){
//说明减小
temp++;
}else{
//否则
if(temp > d_count){
d_count = temp;//更新
}
temp = 1;
}
}
if(temp > d_count){
//如果一直到最后，可能缺少一次跟新
d_count = temp;
}
if(i_count >= d_count)
return i_count;
else return d_count;
}
}
397. Longest Continuous Increasing Subsequence
标签：app 个数 ++ HERE info find write 否则 line
原文地址：https://www.cnblogs.com/phdeblog/p/9313335.html
