數組元素限定條件下的最大距離

//對於數組a[n],找出max值,max值條件如下://0<=i<j; 存在p=a[i],q=a[j], p<=q && p與q之間不存在其他值(a[k])//max = abs(q-p)//如果不存在max max=-2//如果max > 100000 max=-1int GetNum(int a[], int n){int i,j,t;int *b = new int[n];int max = -2;bool flag = true;for(i=0; i<n; i++){b[i] = i;}for(i=0; i<n-1 && flag; i++){flag = false;for(j=0; j<n-1-i; j++){if(a[j] > a[j+1]){t = a[j];a[j] = a[j+1];a[j+1] = t;t = b[j];b[j] = b[j+1];b[j+1] = t;flag = true;}}}for(i=0; i<n-1; i++){if(b[i] < b[i+1]){t = a[i+1] - a[i];t = t>0? t:(-1)*t;if(t > 100000)t = -1;if(t > max)max = t;}}//debugfor(i=0; i<n; i++){printf("%d ",a[i]);}printf("\n");for(i=0; i<n; i++){printf("%d ",b[i]);}delete []b;return max;}