Subarray Sum Closest(lintcode 139)
Description
Given an integer array, find a subarray with sum closest to zero. Return the indexes of the first number and last number.
Example
Given [-3, 1, 1, -3, 5], return [0, 2], [1, 3], [1, 1], [2, 2] or [0, 4].
Interface
public class Solution {
/**
* @param nums: A list of integers
* @return: A list of integers includes the index of the first number
* and the index of the last number
*/
public int[] subarraySumClosest(int[] nums) {
// write your code here
}
}
Solution
public class Solution {
/**
* @param nums: A list of integers
* @return: A list of integers includes the index of the first number
* and the index of the last number
*/
public int[] subarraySumClosest(int[] nums) {
// write your code here
int[] res = new int[2];
if (nums == null || nums.length == 0) {
return res;
}
if (nums.length == 1) {
res[0] = res[1] = 0;
return res;
}
Pair[] sums = new Pair[nums.length + 1];
sums[0] = new Pair(0, 0);
int prev = 0;
for (int i = 1; i <= nums.length; ++i) {
sums[i] = new Pair(prev + nums[i - 1], i);
prev = sums[i].sum;
}
Arrays.sort(sums, new Comparator<Pair>() {
@Override
public int compare(Pair a, Pair b) {
return a.sum - b.sum;
}
});
int min = Integer.MAX_VALUE;
for (int i = 1; i <= nums.length; ++i) {
if (min > sums[i].sum - sums[i - 1].sum) {
min = sums[i].sum - sums[i - 1].sum;
int[] temp = new int[]{sums[i].index - 1, sums[i - 1].index - 1};
Arrays.sort(temp);
res[0] = temp[0] + 1;
res[1] = temp[1];
}
}
return res;
}
}
class Pair {
int sum;
int index;
public Pair(int sum, int index) {
this.sum = sum;
this.index = index;
}
}