From 600120d1b523c56bc1c2b9bb261fb55a9f761df7 Mon Sep 17 00:00:00 2001
From: Ragini Singh <87365673+raginiisingh@users.noreply.github.com>
Date: Sun, 2 Oct 2022 23:04:38 +0530
Subject: [PATCH] Added the solution to the question in Java.

---
 leetCode Solutions/Q8_atoi/Q8_atoi.java | 25 +++++++++++++++++++++++++
 1 file changed, 25 insertions(+)
 create mode 100644 leetCode Solutions/Q8_atoi/Q8_atoi.java

diff --git a/leetCode Solutions/Q8_atoi/Q8_atoi.java b/leetCode Solutions/Q8_atoi/Q8_atoi.java
new file mode 100644
index 0000000..4f87fbb
--- /dev/null
+++ b/leetCode Solutions/Q8_atoi/Q8_atoi.java	
@@ -0,0 +1,25 @@
+/* This solution uses the concept and strings and the functions concerning it. */
+
+Class answer{
+  public int atoi(String s){
+    if(s==null) return 0;     //in case the String is empty
+    s=s.trim();  //removes leading spaces
+    if(s.length()==0) return 0;
+    int sign=-1;
+    long ans=0;   //taking long since integer might exceed the max/min
+    int max=Integer.MAX_VALUE;
+    int min=Integer.MIN_VALUE;
+    if(s.charAt(0)=='-') sign = -1; 
+        int i = (s.charAt(0)=='+' || s.charAt(0)=='-') ? 1 : 0; //skips an index if a sign is there.
+        while(i<s.length()){
+            if(!Character.isDigit(s.charAt(i))) break;    //returns 0 if first index is not a digit and breaks out of the loop.
+            ans = ans * 10 + (s.charAt(i)-'0'); 
+            if(sign == -1 && ans*-1 < MIN) return MIN; //if ans exceeds min then return min according to the question.
+            if(sign == 1 && ans > MAX) return MAX;  //if ans exceeds max then return max according to the question.
+            i++;
+        }
+        return (int)ans*sign;
+  }
+}
+    
+