[Leetcode] 数组题汇总

目录

152. 乘积最大子数组

209. 长度最小的子数组

152. 乘积最大子数组

public class MaxProduct152 {
    public int maxProduct(int[] nums) {
        int  n = nums.length;
        int[] dp = new int[n];
        dp[0]=nums[0];
        int result  = nums[0], nmax = nums[0], nmin = nums[0];
        for(int i = 1;i<n;i++){
            int temp1 = nmax*nums[i], temp2 = nmin*nums[i];
            nmax = Math.max(Math.max(temp1, temp2), nums[i]);
            nmin = Math.min(Math.min(temp1, temp2), nums[i]);
            result = Math.max(nmax, result);

        }
        return result;
    }

    public static void main(String[] args) {
        MaxProduct152 solution = new MaxProduct152();
        int[] nums = new int[]{2,3,-2,4};
        System.out.println(solution.maxProduct(nums));
    }
}

209. 长度最小的子数组

public class MinSubArrayLen {
 
    // 209. 长度最小的子数组
    public int minSubArrayLen(int target, int[] nums) {
        int sum = 0;
        LinkedList<Integer> que = new LinkedList<>();
        int minlen = nums.length + 1;
 
        for (int num : nums) {
            sum += num;
            que.addLast(num);
            while (sum >= target) {
                minlen = Math.min(que.size(), minlen);
                int value = que.removeFirst();
                sum = sum - value;
            }
        }
        return minlen == nums.length + 1 ? 0 : minlen;
 
    }
 
 
    public static void main(String[] args) {
        int[] nums = new int[]{1, 1, 1, 1, 1, 1, 1, 1};
        MinSubArrayLen solution = new MinSubArrayLen();
        System.out.println(solution.minSubArrayLen(11, nums));
    }
}