分享下头条第一题解法，同时求问第四题这类题该如何做

vector < int >res( 2 ,- 1 );

if (nums. size ()<= 2 ){

res[ 0 ]=- 1 ;

res[ 1 ]=- 1 ;

} else {

vector < int >dpDecrease(nums. size (), 1 ); // 序列到当前位的最长递减子串长度

vector < int >dpIncrease(nums. size (), 1 ); // 序列到当前位的最长递增子串长度

for ( int i= 1 ;i<nums. size ();++i){

if (nums[ i ]<nums[ i - 1 ]){

dpDecrease[ i ] = dpDecrease[ i - 1 ]+ 1 ;

dpIncrease[ i ] = 1 ;

} else {

dpIncrease[ i ] = dpIncrease[ i - 1 ]+ 1 ;

dpDecrease[ i ] = 1 ;

}

}

vector < int >len(nums. size (), 0 ); // 序列到当前位的最长 “ 先上升后下降 ” 区间长度

int max = 1 ;

int left = - 1 ,right=- 1 ;

for ( int i= 0 ;i<nums. size ();++i){

if (dpDecrease[ i ]== 1 ){

len[ i ]= 0 ;

} else {

int j = i - dpDecrease[ i ] + 1 ;

if (dpIncrease[ j ] == 1 ){

len[ i ] = 0 ;

} else {

len[ i ] = dpDecrease[ i ]+dpIncrease[ j ]- 1 ;

if (max<len[ i ]){

max = len[ i ];

left = j - dpIncrease[ j ]+ 1 ;

right = i;

}

}

}

}

res[ 0 ] = left;

res[ 1 ] = right;

}

return res;

}