LintCode 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.
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Example
Given [-3, 1, 2, -3, 4], return [0, 2] or [1, 3].
Note
There is at least one subarray that it's sum equals to zero.
卡了一会,需要求一个sum = sigma(a[0]..a[i])的累加和,因为如果某个subarray是0的话,就会出现有两个累加和相同的请求。比如上面例子中的累加和:
-3 -2 0 -3 -1
可以看到有两个-3,也就是说在这两者之间其subarray和为0,才导致了两个位置累加和重叠。还有一种简单的情况就是碰到摸个累加和为0,那么也就是说从数组开始到当前位置的和为0。
class Solution {
public:
/**
* @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
*/
vector<int> subarraySum(vector<int> nums){
// write your code here
int len = nums.size();
vector<int> res;
int sum = 0;
unordered_map<int, int> val2idx;
for (int i = 0; i < len; i++) {
sum += nums[i];
if (sum == 0) {
res = {0, i};
return res;
}
if (val2idx.count(sum) > 0) {
res = {val2idx[sum] + 1, i};
break;
}
val2idx[sum] = i;
}
return res;
}
};
感觉放在easy等级里不合适