leetcode2529--正整数和负整数的最大计数

1. 题意

给定有序数组，求其中正整数和负整数的计数最大值。

正整数和负整数的最大计数

2. 题解

2.1 遍历

直接判断

class Solution {
public:
    int maximumCount(vector<int>& nums) {
        
        int neg = 0;
        int pos = 0;

        for (int num:nums) {
            if (!num)
                continue;
            if (num & 0x80000000 )
                neg++;
            else
                pos++;
        }
        return neg < pos ? pos : neg;
    }
};

2.2 二分查找

二分查找第一个非负值和第一个正值。

class Solution {
public:
    bool check(int l, int r, vector<int> &nums, int target)
    {
        int sum = 0;
        for (int i = l; i < r + 1; ++i)
            sum += nums[i];
        
        return sum >= target;
    }

    int minSubArrayLen(int target, vector<int>& nums) {

        int l = 0;
        int sum = 0;
        int sz = nums.size();
        int ans = INT_MAX;


        vector<int> pre(sz + 1, 0);
        int preSum = 0;
        for (int i = 0;i < sz; ++i) {
            preSum += nums[i];
            pre[i + 1] = preSum;
        }

        
        for (int i = 0;i < sz; ++i) {
            
            int l = i;
            int r = sz;

            while (l < r) {
                int m = ((r - l) >> 1) + l;

               // cout << "m"<< m << "," << "sum" << sum << endl; 
                if ( pre[m + 1] - pre[i] < target)
                    l = m + 1;
                else
                    r = m;

            }
           // printf("%d,%d\n", i, l);
            if ( l < sz){
                ans = min(l - i + 1, ans);
            }
        }


        return ans == INT_MAX ? 0 : ans;
    }
};

使用库函数

class Solution {
public:
    int maximumCount(vector<int>& nums) {
        
        auto non_neg = lower_bound(nums.begin(), nums.end(), 0) - nums.begin();
        auto pos = upper_bound(nums.begin(), nums.end(), 0);


        return max(non_neg, nums.end()- pos);
    }
};