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. Hashmap is also used in this problem. Iterate the array and calculate the su
Maximum Size Subarray Sum Equals k -- LeetCode Given an arraynumsand a target valuek, find the maximum length of a subarray that sums tok. If there isn't one, return 0 instead. Example 1: Givennums=[1, -1, 5, -2, 3],k=3, return4. (because the subarray[1, -1, 5, -2]su...
初始化的目的:如果当前位置和为k,那差值就是0,但是存储的值却是sum,不一定存储了0的值。 classSolution {public:intsubarraySum(vector<int>& nums,intk) { unordered_map<int,int>m; m[0] =1;intsum =0,res =0;for(inti =0;i < nums.size();i++){ sum+=nums[i]; res+= m[sum -k]; ...
LeetCode 325. Maximum Size Subarray Sum Equals k 若谷 追求卓越,成功就会在不经意间追上你。 来自专栏 · 今日事 题目: 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 ...
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and clearly, we need a global variable to keep record of the global maximum. so the code will be: class Solution { public int maxSubArray(int[] nums) { if(nums==null || nums.length == 0) return 0; int[] dp = new int[nums.length]; ...
LeetCode "Maximum Size Subarray Sum Equals k" Nothing special. Point is, how clean can you do it? classSolution {public:intmaxSubArrayLen(vector<int>& nums,intk) {intn =nums.size();//Partial Sumvector<int>asum(n); partial_sum(nums.begin(), nums.end(), asum.begin());intret =...
(vector<int> &a, int l, int r) { if (l == r) { return (Status) {a[l], a[l], a[l], a[l]}; } int m = (l + r) >> 1; Status lSub = get(a, l, m); Status rSub = get(a, m + 1, r); return pushUp(lSub, rSub); } int maxSubArray(vector<int>& nums...
in prefixSum array, find the maxisum size subarray sum equals k// sum of subarray (i, j) = prefixSum [j] - prefixSum[i]Map<Integer,Integer>prefixSumVsIndex=newHashMap<>();prefixSumVsIndex.put(0,-1);intmaxSize=0;intprefixSum=0;for(intj=0;j<nums.length;j++){prefixSum+=nums[...
the contiguous subarray [4,-1,2,1] has the largest sum = 6.中文:主要是给定一个数组,求解数组的子数组中,数组元素和最大的那一个子数组,返回的是最大子数组的和。2. 求解解法一最简单也是最容易想到的思路就是三层循环,对(i,j),i<=j的情况进行遍历,这种情况下的算法复杂度为O($n^3$)。代码如...