二分算法框架

  • 1)最基本的二分搜索
public int binarySearch(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) {
            return mid;
        } else if (nums[mid] > target) {
            right = mid - 1;
        } else if (nums[mid] < target) {
            left = mid + 1;
        }
    }
    return -1;
}
  • 2)寻找左侧边界
//在arr上，找满足>=value的最左位置
//例 [1,1,2,2,2,2,3,3,4]  value=2 则index=2
public int leftBound(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) {
            right = mid - 1;
        } else if (nums[mid] > target) {
            left = mid + 1;
        } else if (nums[mid] < target) {
            right = mid - 1;
        }
    }
    //left = right + 1, left即为要找的值
    if (left > nums.length || nums[left] != target)
        return -1;
    return left;
}
  • 3)寻找右侧边界
public int rightBound(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) {
            right = mid - 1;
        } else if (nums[mid] > target) {
            left = mid + 1;
        } else if (nums[mid] < target) {
            right = mid - 1;
        }
    }
    //left = right + 1, right即为要找的值
    if (right < 0 || nums[right] != target)
        return -1;
    return right;
}
  • 4)局部最小值问题
public int getLessIndex(int[] nums) {
    if (nums == null || nums.length == 0) return -1;
    if (nums.length == 1 || nums[0] < nums[1]) return 0;
    if (nums[nums.length - 1] < nums[nums.length - 2]) return nums.length - 1;

    int left = 0, right = nums.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] > nums[mid - 1]) {
            right = mid - 1;
        } else if (nums[mid] > nums[mid + 1]) {
            left = mid + 1;
        } else {
            return mid;
        }
    }
    return -1;
}

一些二分STL（适用于排序后的数组）

  • lower_bound( begin,end,num)：从数组的begin位置到end-1位置二分查找第一个大于或等于num的数字，找到返回该数字的地址，不存在则返回end。
  • upper_bound( begin,end,num)：从数组的begin位置到end-1位置二分查找第一个大于num的数字，找到返回该数字的地址，不存在则返回end。

例题

const int maxn = 1e5 + 6;
int n, k;
ll a[maxn];

//判断p是否合适
int check(ll p) {
    int i = 0;
    for (int j = 0; j < k; j++) {
        ll s = 0;
        while (s + a[i] <= p) {
            s += a[i];
            i++;
            if (i == n) return n;
        }
    }
    return i;
}

ll go() {
    ll left = 0;
    ll right = 1e9 + 1;
    ll mid;
    //使用二分
    while (left + 1 < right) {
        mid = m(left, right);
        int v = check(mid);
        if (v >= n) right = mid;
        else left = mid;
    }
    return right;
}

int main() {
    cin >> n >> k;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    ll ans = go();
    cout << ans << endl;
}

题目：判断子序列

  • 解法一：双指针
    从左到右遍历一次即可
class Solution {
    public boolean isSubsequence(String s, String t) {
        int m = s.length();
        int n = t.length();
        int i = 0, j = 0;
        while (i < m && j < n) {
            if (s.charAt(i) == t.charAt(j)) {
                i++;
                j++;
            } else {
                j++;
            }
        }
        return i == m;
    }
}
  • 解法二：二分算法

用hash表遍历一次总字符串，记录a~z出现的index，要求s[i]比s[i-1]的对应t中的index大即可。

class Solution {
    public boolean isSubsequence(String s, String t) {
        HashMap<Character, ArrayList<Integer>> map = new HashMap<>();
        for (char c = 'a'; c <= 'z'; c++) {
            map.put(c, new ArrayList<>());
        }
        for (int i = 0; i < t.length(); i++) {
            map.get(t.charAt(i)).add(i);
        }
        int index = -1;
        for (char c : s.toCharArray()) {
            ArrayList<Integer> tmp = map.get(c);
            int left = 0, right = tmp.size();
            int tmp_index = -1;
            while (left < right) {
                int mid = left + ((right - left) >> 1);
                if (tmp.get(mid) > index) {
                    tmp_index = tmp.get(mid);
                    right = mid;
                } else {
                    left = mid + 1;
                }
            }
            if (tmp_index != -1) {
                index = tmp_index;
            } else {
                return false;
            }
        }
        return true;
    }
}

题目：元素和小于等于阈值的正方形的最大边长

给你一个大小为 m x n 的矩阵 mat 和一个整数阈值 threshold。

请你返回元素总和小于或等于阈值的正方形区域的最大边长；如果没有这样的正方形区域，则返回 0 。

思路

  1. 利用二维前缀和表示出数组的和dp[i][j] = mat[i - 1][j - 1] + dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1],这里dp[i][j]表示mat[i-1][j-1]的前缀和，之后可以使用dp[i][j] - dp[i - k][j] - dp[i][j - k] + dp[i - k][j - k]来表示以dp[i][j]为结尾、k为边长的正方形的和
  2. 使用一个方法来计算边长k是否符合要求（边长为k的正方形元素总和小于或等于阈值）
  3. 利用二分得出[0,min(m,n)]中符合要求的最大边长
    • 若mid符合要求，则left=mid+1
    • 若mid不符合要求，则right=mid-1

代码

class Solution {
public:
    static const int maxn = 302;
    int dp[maxn][maxn], m, n;

    int maxSideLength(vector<vector<int>> &mat, int threshold) {
        m = mat.size(), n = mat[0].size();
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                dp[i][j] = mat[i - 1][j - 1] + dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1];
            }
        }
        int left = 0, right = min(m, n), ans = 0;
        while (left <= right) {
            int mid = left + ((right - left) >> 1);
            if (help(mid, threshold)) {
                ans = mid;
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return ans;
    }

    bool help(int k, int shold) {
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (i - k < 0 || j - k < 0) {
                    continue;
                }
                if (dp[i][j] - dp[i - k][j] - dp[i][j - k] + dp[i - k][j - k] <= shold) {
                    return true;
                }
            }
        }
        return false;
    }
};

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