标签：new inpu find highlight running get numbers boolean 多个
Given an array of integers and an integer k,
you need to find the total number of continuous subarrays whose sum equals to k. Example 1: Input:nums = [1,1,1], k = 2 Output: 2 Note: The length of the array is in range [1, 20,000]. The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].
subarray sum 问题常用hashmap, 存count 值和坐标, 动归的感觉啊
public int subarraySum(int[] nums, int k) {
int sum = 0, result = 0;
Map<Integer, Integer> preSum = new HashMap<>();
preSum.put(0, 1);
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
if (preSum.containsKey(sum - k)) {
result += preSum.get(sum - k);
}
// 当加着加着出现两个一样的sum时, 要在他的value上加1, 因为可以有多个连续的串
preSum.put(sum, preSum.getOrDefault(sum, 0) + 1);
}
return result;
}
要用 preSum.put(0, 1); 是得result 加的值可以来自map中的多个.
不能 if (sum == k) {
result++;
}
因为:
public boolean checkSubarraySum(int[] nums, int k) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>(){{put(0,-1);}};;
int runningSum = 0;
for (int i=0;i<nums.length;i++) {
runningSum += nums[i];
if (k != 0) runningSum %= k;
Integer prev = map.get(runningSum);
if (prev != null) {
if (i - prev > 1) return true;
}
else map.put(runningSum, i);
}
return false;
}
标签：new inpu find highlight running get numbers boolean 多个
原文地址：http://www.cnblogs.com/apanda009/p/7282555.html
