**[Lintcode]Subarray Sum Closest最接近零的子数组和
2016-11-20 16:16
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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.
Example
Given
return
分析:注意,线段树无法解决此类问题。可以考虑用数组res[i]储存0~i的和。此后根据和排序。然后寻找和最接近的两个元素。如果有两个元素和接近,则代表在原数组中,这两个元素所代表的数组段的差异部分的和接近0. 最后,为了可以返回index,这里可以使用一个数据结构来一起保存sum和index。
利用sum排序,利用index返回数组坐标。
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 class Data {
int index, sum;
}
public int[] subarraySumClosest(int[] nums) {
if(nums.length == 0) return new int[0];
if(nums.length == 1) return new int[]{0, 0};
Data[] res = new Data[nums.length];
int sum = 0;
for(int i = 0; i < nums.length; i++) {
sum += nums[i];
res[i] = new Data();
res[i].index = i;
res[i].sum = sum;
}
Arrays.sort(res, new Comparator<Data>() {
public int compare(Data d1, Data d2) {
return d1.sum - d2.sum;
}
});
int min = Integer.MAX_VALUE, left = -1, right = -1;
for(int i = 0; i < res.length - 1; i++) {
if(Math.abs(res[i].sum - res[i + 1].sum) < min) {
min = Math.abs(res[i].sum - res[i + 1].sum);
//left = left + 1, right = right
left = res[i].index < res[i + 1].index ? res[i].index + 1 : res[i + 1].index + 1;
right = res[i].index < res[i + 1].index ? res[i + 1].index : res[i].index;
}
}
return new int[]{left, right};
}
}
Example
Given
[-3, 1, 1, -3, 5],
return
[0, 2],
[1, 3],
[1, 1],
[2, 2]or
[0, 4].
分析:注意,线段树无法解决此类问题。可以考虑用数组res[i]储存0~i的和。此后根据和排序。然后寻找和最接近的两个元素。如果有两个元素和接近,则代表在原数组中,这两个元素所代表的数组段的差异部分的和接近0. 最后,为了可以返回index,这里可以使用一个数据结构来一起保存sum和index。
利用sum排序,利用index返回数组坐标。
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 class Data {
int index, sum;
}
public int[] subarraySumClosest(int[] nums) {
if(nums.length == 0) return new int[0];
if(nums.length == 1) return new int[]{0, 0};
Data[] res = new Data[nums.length];
int sum = 0;
for(int i = 0; i < nums.length; i++) {
sum += nums[i];
res[i] = new Data();
res[i].index = i;
res[i].sum = sum;
}
Arrays.sort(res, new Comparator<Data>() {
public int compare(Data d1, Data d2) {
return d1.sum - d2.sum;
}
});
int min = Integer.MAX_VALUE, left = -1, right = -1;
for(int i = 0; i < res.length - 1; i++) {
if(Math.abs(res[i].sum - res[i + 1].sum) < min) {
min = Math.abs(res[i].sum - res[i + 1].sum);
//left = left + 1, right = right
left = res[i].index < res[i + 1].index ? res[i].index + 1 : res[i + 1].index + 1;
right = res[i].index < res[i + 1].index ? res[i + 1].index : res[i].index;
}
}
return new int[]{left, right};
}
}
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