[LeetCode#1] Two Sum

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The Problem:

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

Analysis: 

The idea of the algorithm is classic, we need to use the search window and invariant to understand the idea behind it. 

1. search window 

front ......... end

the search window is  num[front, end]. 

2. the invariant <The right answers must be in the search window>

we only to make sure that the right elements are always in the search window when we make decision about moving pointer, we could finally reach the right answer. 

Note : assume the num array has already been sorted.

iff num[front] + num[end] < target, we move front : front ++.

All combination of num[i] and num[j], meeting the condition num[i] + num[j] > num[front] + num[end] is the search window. the answer must in it too!!!  

iff num[front] + num[end] > target, we move end : end --. 

The invariant: the right answer is always in the search window. (keep closing by moving pointer)

proof:

iff num[front] + num[end] < target,

we move end to end‘: end --. num[front] + num[end‘] < target, since num[end‘] <= num[end].

The right solution may not be contained in the search window. 

Pitfall:

Since the question ask us to return the index, so we need to research the answer‘s index at its original array. (it means we should copy the whole array to make the sorting.)

My solution: 

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        
        int len = numbers.length;
        int[] sorted = new int[len];
        int[] obj = new int[2]; //used to contain value that meet obj[0] + obj[1] = target 
        int[] ret = new int[2]; //used to contain the index of obj[0] and obj[1] in original array
        int index_front;
        int index_end;
      
        System.arraycopy(numbers, 0, sorted, 0, len); // must use a copy to sort, at last, we need to finde the original index
        Arrays.sort(sorted);
        
        int front = 0;
        int end = len - 1;
        
        while (front < end) {  //take care of the index conversion!
            if (sorted[front] + sorted[end] == target) {
                break;
            } else if (sorted[front] + sorted[end] < target ) {
                front ++;
            } else {
                end --; 
            }
        }
        
        obj[0] = sorted[front];
        obj[1] = sorted[end];
        ret = find_indexes(numbers, obj);
        
        ret[0] ++;
        ret[1] ++;
        Arrays.sort(ret);
        return ret;
    }
    
    static int[] find_indexes(int[] numbers, int[] obj) { 
        int[] ret = new int[2];
        ret[0] = -1; //must set to -1, cause the following check condition 
        ret[1] = -1; 
        for (int i = 0; i < numbers.length; i++) { //must check ret[0] == -1, otherwise, ret[1] would never be assigned
                if (numbers[i] == obj[0] && ret[1] != i && ret[0] == -1) { // avoid to count the same element twice
                    ret[0] = i;
                }
                if (numbers[i] == obj[1] && ret[0] != i && ret[1] == -1) { 
                    ret[1] = i; 
                }
        }
        return ret;
    }
}

note: when we search for two distinct elements in the same array, we must make sure they were assigned with different indexes(elements).

A tricky skill to achieve is to check if the element has already been assigned, by checking the ans[m] == i . 

        ret[0] = -1; //must set to -1, cause the following check condition 
        ret[1] = -1; 
        for (int i = 0; i < numbers.length; i++) { //must check ret[0] == -1, otherwise, ret[1] would never be assigned
                if (numbers[i] == obj[0] && ret[1] != i && ret[0] == -1) { // avoid to count the same element twice
                    ret[0] = i;
                }
                if (numbers[i] == obj[1] && ret[0] != i && ret[1] == -1) { 
                    ret[1] = i; 
                }
        }

numbers[i] == obj[0]: guarantee the element is the right element we searching for.

ret[1] != i : guarantee the element was not assigned to others. 

ret[0] == -1 : guarantee the elements‘ with the same vlue are not assgined to a single variable twice. 
take care: (The elements with the same value!)(-1 is great when search in array)

[LeetCode#1] Two Sum

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原文地址：http://www.cnblogs.com/airwindow/p/4192759.html