Java二分查找算法与数组处理的应用实例

class Solution {
    public int specialArray(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        int left = 0, right = n;
        while (left <= right) {
            int x = (left + right) >> 1;
            int idx = binarySearch(nums, x); // nums中第一个大于等于x的元素位置
            if (x == n - idx) {
                return x;
            } else if (x < n - idx) { // 大于等于x的元素太多了，所以下一轮搜索要增大x的取值范围
                left = x + 1;
            } else { // 反之，减少x的取值范围
                right = x - 1;
            }
        }
        return -1;
    }
    private static int binarySearch(int[] nums, int x) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = (left + right) >> 1;
            int val = nums[mid];
            if (val >= x) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

class Solution {
    public int shipWithinDays(int[] weights, int days) {
        // 确定二分查找左右边界
        int left = Arrays.stream(weights).max().getAsInt(), right = Arrays.stream(weights).sum();
        while (left < right) {
            int mid = (left + right) / 2;
            // need 为需要运送的天数
            // cur 为当前这一天已经运送的包裹重量之和
            int need = 1, cur = 0;
            for (int weight : weights) {
                if (cur + weight > mid) {
                    ++need;
                    cur = 0;
                }
                cur += weight;
            }
            if (need <= days) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

class Solution {
    public int[] successfulPairs(int[] spells, int[] potions, long success) {
        int[] ans = new int[spells.length];
        Arrays.sort(potions);
        for (int i = 0; i < spells.length; i++) {
            int power = spells[i];
            int left = 0;
            int right = potions.length - 1;
            while (left <= right) {
                int mid = left + (right - left) / 2;
                if (1l * power * potions[mid] < success) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            }
            ans[i] = potions.length - left;
        }
        return ans;
    }
}