821. Shortest Distance to a Character

Question

821. Shortest Distance to a Character

Solution

思路：遍历字符串S,遇到与字符C相等就分别向左/右计算其他字符与该字符的距离，如果其他字符就是C或与目标字符的距离更小的话遍历就终止。

Java实现：

public int[] shortestToChar(String S, char C) {
    int[] store = new int[S.length()];
    for (int i = 0; i < S.length(); i++) {
        if (S.charAt(i) == C) {
            store[i] = 0;
            left(S, C, store, i);
            right(S, C, store, i);
        }
    }
    return store;
}

void left(String S, char C, int[] store, int target) {
    for (int i = target - 1; i >= 0; i--) {
        if (S.charAt(i) == C) break;
        if (store[i] == 0 || store[i] > target - i) store[i] = target - i;
    }
}

void right(String S, char C, int[] store, int target) {
    for (int i = target + 1; i < S.length(); i++) {
        if (S.charAt(i) == C) break;
        if (store[i] == 0 || store[i] > i - target) store[i] = i - target;
    }
}