// Given an array nums and a target value k, find the maximum length of a subarray that sums to k.
//If there isn't one, return 0 instead.
// Note:
// The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range.
// Example 1:
// Given nums = [1, -1, 5, -2, 3], k = 3,
// return 4. (because the subarray [1, -1, 5, -2] sums to 3 and is the longest)
// Example 2:
// Given nums = [-2, -1, 2, 1], k = 1,
// return 2. (because the subarray [-1, 2] sums to 1 and is the longest)
// Follow Up:
// Can you do it in O(n) time?
public class MaximumSizeSubarraySumEqualsK {
public int maxSubArrayLen(int[] nums, int k) {
if(nums.length == 0) {
return 0;
}
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int maxLength = 0;
int total = 0;
map.put(0, -1);
for(int i = 0; i < nums.length; i++) {
total += nums[i];
if(map.containsKey(total - k)) {
maxLength = Math.max(maxLength, i - map.get(total - k));
}
if(!map.containsKey(total)) {
map.put(total, i);
}
}
return maxLength;
}
}
//If there isn't one, return 0 instead.
// Note:
// The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range.
// Example 1:
// Given nums = [1, -1, 5, -2, 3], k = 3,
// return 4. (because the subarray [1, -1, 5, -2] sums to 3 and is the longest)
// Example 2:
// Given nums = [-2, -1, 2, 1], k = 1,
// return 2. (because the subarray [-1, 2] sums to 1 and is the longest)
// Follow Up:
// Can you do it in O(n) time?
public class MaximumSizeSubarraySumEqualsK {
public int maxSubArrayLen(int[] nums, int k) {
if(nums.length == 0) {
return 0;
}
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int maxLength = 0;
int total = 0;
map.put(0, -1);
for(int i = 0; i < nums.length; i++) {
total += nums[i];
if(map.containsKey(total - k)) {
maxLength = Math.max(maxLength, i - map.get(total - k));
}
if(!map.containsKey(total)) {
map.put(total, i);
}
}
return maxLength;
}
}
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