the contiguous subarray[4,−1,2,1]has the largest sum =6. click to show more practice. Subscribeto see which companies asked this question 1publicclassSolution {2publicintmaxSubArray(int[] nums) {3if(nums ==null)return0;4if(nums.length == 1)returnnums[0];5intcurmax = 0;6intans ...
classSolution {publicintmaxSubArray(int[] nums) {intmaxSum = Integer.MIN_VALUE;//注意有负数的时候,不能初始化为0intcurrentSum =Integer.MIN_VALUE;for(inti = 0; i < nums.length; i++){if(currentSum < 0) currentSum =nums[i];elsecurrentSum +=nums[i];if(currentSum > maxSum) maxSum ...
class Solution { public int maxSubArray(int[] nums) { int pre = 0, maxAns = nums[0]; for (int x : nums) { pre = Math.max(pre + x, x); maxAns = Math.max(maxAns, pre); } return maxAns; } } 方法2:分治 class Solution { public class Status { public int lSum, rSum, ...
Follow up: If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle. class Solution { public int maxSubArray(int[] nums) { int maxSum = Integer.MIN_VALUE; int curSum = 0; int left = 0, right = 0; while ...
// add 53. Maximum Subarray class Solution { public: int maxSubArray(vector<int>& nums) { int ret = INT_MIN; int temp = 0; for (int i = 0; i < nums.size();i++) { temp += nums[i]; if (temp>ret) // //两个if的先后顺序 ...
importjava.util.stream.IntStream; /** * 最大连续子数组的乘积 * * Find the contiguous subarray within an array (containing at least one number) which has the largest product. * * For example, given the array [2,3,-2,4], * the contiguous...
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好吧,还是习惯了Java的三元运算符的写法。 有了递推公式的话,那么代码就很简单了 classSolution(object):defmaxSubArray(self,nums):""" :type nums: List[int] :rtype: int """iflen(nums)<1:return0ret=nums[0]dp=[]dp.append(nums[0])foriinrange(1,len(nums)):ifdp[i-1]<0:dp.append(num...
Runtime: 1 ms, faster than 93.39% of Java online submissions for Maximum Product Subarray. classSolution{publicintmaxProduct(int[]nums){intn=nums.length;intmax=nums[0];intmin=nums[0];intres=max;for(inti=1;i<n;i++){inttMax=Math.max(nums[i]*max,nums[i]*min);inttMin=Math.min(num...
调用helper 函数 分别得到max Subarray 从0到i 和从 size - 1 到 i + 1的max subarray, 让后遍历扫描线 题解代码 java public class Solution { /* * @param nums: A list of integers * @return: An integer denotes the sum of max two non-overlapping subarrays */ private int maxSubArrayLeft(...