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Given an integer array, find a subarray with sum closest to zero. Return the indexes of the first number and last number.
Given [-3, 1, 1, -3, 5], return [0, 2], [1, 3], [1, 1], [2, 2] or [0, 4]
O(nlogn) time
Analysis:
s[i] = nums[0]+....nums[i], also record the index i into s[i]. Sort array s, and the minimum difference between two consecutive element, is the the subarray.
Solution:
1 class Element implements Comparable<Element>{ 2 int val; 3 int index; 4 public Element(int v, int i){ 5 val = v; 6 index = i; 7 } 8 9 public int compareTo(Element other){ 10 return this.val - other.val; 11 } 12 13 public int getIndex(){ 14 return index; 15 } 16 17 public int getValue(){ 18 return val; 19 } 20 } 21 22 public class Solution { 23 /** 24 * @param nums: A list of integers 25 * @return: A list of integers includes the index of the first number 26 * and the index of the last number 27 */ 28 public ArrayList<Integer> subarraySumClosest(int[] nums) { 29 ArrayList<Integer> res = new ArrayList<Integer>(); 30 if (nums.length==0) return res; 31 32 Element[] sums = new Element[nums.length+1]; 33 sums[0] = new Element(0,-1); 34 int sum = 0; 35 for (int i=0;i<nums.length;i++){ 36 sum += nums[i]; 37 sums[i+1] = new Element(sum,i); 38 } 39 40 Arrays.sort(sums); 41 int min = Math.abs(sums[0].getValue() - sums[1].getValue()); 42 int start = Math.min(sums[0].getIndex(), sums[1].getIndex())+1; 43 int end = Math.max(sums[0].getIndex(), sums[1].getIndex()); 44 for (int i=1;i<nums.length;i++){ 45 int diff = Math.abs(sums[i].getValue() - sums[i+1].getValue()); 46 if (diff<min){ 47 min = diff; 48 start = Math.min(sums[i].getIndex(), sums[i+1].getIndex())+1; 49 end = Math.max(sums[i].getIndex(), sums[i+1].getIndex()); 50 } 51 } 52 53 res.add(start); 54 res.add(end); 55 return res; 56 } 57 }
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原文地址:http://www.cnblogs.com/lishiblog/p/4187961.html