Lintcode: Subarray Sum 解题报告

Subarray Sum

原题链接：http://lintcode.com/zh-cn/problem/subarray-sum/#

Given an integer array, find a subarray where the sum of numbers is zero. Your code should return the index of the first number and the index of the last number.

样例

Given [-3, 1, 2, -3, 4], return [0, 2] or [1, 3].

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SOLUTION 1:

我们有一个O(N)的解法。使用Map 来记录index, sum的值。当遇到两个index的sum相同时，表示从index1+1到index2是一个解。

注意：添加一个index = -1作为虚拟节点。这样我们才可以记录index1 = 0的解。

空间复杂度：O(N)

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 ArrayList<Integer> subarraySum(int[] nums) {
        // write your code here
        
        int len = nums.length;
        
        ArrayList<Integer> ret = new ArrayList<Integer>();
        
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        
        // We set the index -1 sum to be 0 to let us more convient to count.
        map.put(0, -1);
        
        int sum = 0;
        for (int i = 0; i < len; i++) {
            sum += nums[i];
            
            if (map.containsKey(sum)) {
                // For example: 
                //        -3  1  2 -3 4
                // SUM: 0 -3 -2  0 -3 1
                // then we got the solution is : 0 - 2
                ret.add(map.get(sum) + 1);
                ret.add(i);
                return ret;
            }
            
            // Store the key:value of sum:index.
            map.put(sum, i);
        }
        
        return ret;
    }
}

 

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/lintcode/array/SubarraySum.java