Question:
Given two arrays of ints sorted in increasing order, outer and inner, return true if all of the numbers in inner appear in outer. The best solution makes only a single "linear" pass of both arrays, taking advantage of the fact that both arrays are already in sorted order.
linearIn({1, 2, 4, 6}, {2, 4}) → true
linearIn({1, 2, 4, 6}, {2, 3, 4}) → false
linearIn({1, 2, 4, 4, 6}, {2, 4}) → true
Source:
public boolean linearIn(int[] outer, int[] inner) {boolean b[]=new boolean[inner.length];boolean b1=true;for(int i=0;i<inner.length;i++){for(int j=0 ; j<outer.length;j++){if(inner[i]==outer[j])b[i]=true;}}for(boolean b2:b)b1 &= b2;return b1;}
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fish chuchu · 616 weeks ago
public boolean canBalance(int[] nums) {
int totalSum = 0;
int parSum=0;
for (int i=0;i<nums.length;i++)
{
totalSum = totalSum+nums[i];
}
for(int j=0;j<nums.length;j++)
{
parSum=parSum+nums[j];
if(parSum==totalSum-parSum)
{
return true;
}
}
return false;
}
fish chuchu · 616 weeks ago
public boolean linearIn(int[] outer, int[] inner) {
int count =0;
for (int i=0;i<inner.length;i++)
{
for (int j=0;j<outer.length;j++)
{
if(inner [i]==outer[j])
{
count++;
break;
}
}
}
if (count==inner.length)
{
return true;
}
return false;
}