java - keep getting null pointer exception in linked list -

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इस सवाल का पहले से ही एक उत्तर है: < / P>

  • 12 जवाब

मैं एक को हटाने की कोशिश कर रहा हूँ लिंक की गई सूची से विशेष स्ट्रिंग, लेकिन मैं रिक्त सूचक अपवाद त्रुटि प्राप्त कर रहा हूँ! किसी भी मदद के दोस्तों? 

  सार्वजनिक शून्य हटाना (स्ट्रिंग deleteName) {LinkedListNode current = head; अगर (सिर == रिक्त) {System.out.println ("सूची से हटाने के लिए कुछ भी नहीं! डेटा डालें।"); } // सिर के लिए और अगर (वर्तमान .getName () == deleteName) {head = current.getNext (); } // tail for if else (findTail ()। GetName () == deleteName) {जबकि (current.getNext)। GetNext ()! = नल) {current = current.getNext (); } Current.setNext (नल); } Else {while (current.getNext)। GetName ()! = DeleteName) {current = current.getNext (); } Current.setNext (वर्तमान .getext ()। GetNext ()); } System.out.println (""); System.out.println (deleteName + "सूची से हटा दिया गया है।"); }

कुछ संपादन किया था अब काम करता है! तारों की तुलना करने के लिए == का उपयोग न करें।

<पूर्व> सार्वजनिक शून्य हटाना (स्ट्रिंग deleteName) {लिंक्डलिस्टनोड चालू = सिर; अगर (सिर == रिक्त) {System.out.println ("सूची से हटाने के लिए कुछ भी नहीं! डेटा डालें।"); } // सिर के लिए और अगर (current.getName ()। बराबर (deleteName)) {head = current.getNext (); } // tail for else if (findTail ()। GetName ()। बराबर (deleteName)) {जबकि (current.getNext)। GetNext ()! = नल) {current = current.getNext (); } Current.setNext (नल); } Else {while (! Current.getNext ()। GetName ()। बराबर (deleteName)) {current = current.getNext (); } Current.setNext (वर्तमान .getext ()। GetNext ()); } System.out.println (""); System.out.println (deleteName + "सूची से हटा दिया गया है।"); } // हटाए जाने के अंत ()

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